Plane  j  rigonometry 


JE 


Edward  Bright 


PLANE    TRIGONOMETRY 


BY 


R.    D.    BOHANNAN 

PROFESSOR   OF   MATHEMATICS   IN    THE  OHIO    STATE    UNIVERSITY 


■oo^c 


Boston 

ALLYN    AND    BACON 
1904 


-31* 


COPYRIGHT,  1903,  BY 
R.   D.   BOHANNAN. 


NortoooU  $rfgg 

J.  S.  Cushing  &  Co.  —  Berwick  &  Smith  Co. 

Norwood,  Mass.,  U.S.A. 


L3*c 


PREFACE. 

Some  of  the  features  of  this  book  are : 

(1)  Abandonment  of  the  "  Academic  Triangle." 

The  "  academic  triangle "  is  one,  certain  of  whose  parts  are  known 
exactly  and  whose  remaining  parts  are  calculable  free  from  all  errors 
except  such  as  are  due  to  the  "  place  "  of  the  tables  used.  In  the  academic 
triangle  one  side  may  show  one  figure  and  another  seven  figures,  and 
the  angles  may  show  any  readings  the  caprice  of  the  maker  may  suggest. 

(2)  Replacement  of  the  " Academic  Triangle"  by  triangles 
having  variety  of  form,  but  always  such  form  as  they  should 
have  if  they  had  been  actually  measured  in  the  field: 

(i)  Showing  in  all  measured  lines  of  a  diagram  the  same 
number  of  significant  figures,  thus  indicating  about  the  same 
care  in  each  measurement. 

(ii)  Giving  in  all  calculated  lines  of  a  diagram  not  more 
significant  figures  than  in  its  measured  lines,  since  such 
extra  figures  are  misleading  and  of  no  value. 

(iii)  Letting  the  angles  of  a  diagram  show  such  read- 
ings as  exhibit  care  in  angle- reading  in  keeping  with  that 
indicated  in  line-measurement. 

While  the  student  is  learning  to  calculate,  he  might  as  well  learn  at 
the  same  time  what  form  intelligent  measurement  should  take  and  what 
reliability  calculated  results  have  as  related  to  the  data. 

(3)  What  place  tables  to  use  to  suit  the  data  and  how  to  cut 
a  large  table  to  suit  short  data. 

(4)  How  checks  should  check  to  suit  the  data. 

(5)  The  fundamental  principles  of  calculation  with  approxi- 
mate data  (Chapter  II.). 

(6)  The  suggestion  of  simple  laboratory  exercises  in  connec- 
tion with  fundamental  things. 

iii 


797953 


iv  PREFACE. 

(7)  Exercises  from  related  topics,  as  Physics,  Analytical  Geome- 
try, Surveying,  etc. 

(8)  Graphs,  using  rectangular  and  polar  coordinate  paper. 

(9)  Geometric  treatment  of  DeMoivre's  Theorem. 

(10)  Order  of  treatment  of  the  trigonometric  functions. 

When  many  different  ideas  are  presented  simultaneously  to  the  mind 
the  result  is  confusion.  When  the  same  idea  is  presented  again  and 
again,  with  a  suitable  interval  between  presentations,  its  difficulties  dis- 
appear. Most  of  the  difficulties  connected  with  any  trigonometric  func- 
tion are  those  of  every  other  such  function.  I  have  ventured,  therefore, 
to  present  the  subjects  in  the  following  order  : 

(i)  The  sine,  —  its  inverse  and  reciprocal, 
(ii)  The  cosine,  —  its  inverse  and  reciprocal, 
(iii)  The  sine  family  and  cosine  family  in  union, 
(iv)  The  tangent,  —  its  inverse  and  reciprocal, 
(v)  All  the  functions  in  union, 
(vi)  The  quantity  ^/  —  1  in  trigonometry. 

I  am  indebted  to  Dr.  Coddington  and  Dr.  Kuhn  of  the  Ohio 
State  University,  and  to  Dr.  J.  W.  Young  of  the  Northwestern 
University,  for  valuable  suggestions.  Dr.  Coddington  has  tested 
all  the  exercises. 

R.  D.  BOHANNAN. 

Columbus,  Ohio, 
May,  1904. 


CONTENTS. 

CHAPTER  PAGE 

I.    Elementary  Discussion  of  Logarithms  —  How  to  use  Tables 

and  what  Place  Tables  to  use 1 

II.    Calculation  Vices  and  Devices 32 

III.  Angles  and  Angle-units 60 

IV.  Construction  of  Angles  and  of  Straight-line  Diagrams  to 

Scale,  and  the  Measurement  of  Angles     ....  77 
V.     The  Sine,  Anti-sine,  Reciprocal  Sine  (Cosecant),  and  Co- 
versed  Sine  of  an  Angle 82 

VI.     The  Cosine,  Inverse   Cosine,  Reciprocal  Cosine  (Secant), 

and  Versed  Sine  of  an  Angle 192 

VII.     The  Sine  and  the  Cosine  in  Union 237 

VIII.     The  Tangent  and  Reciprocal  Tangent  (Cotangent)  of  Angles  297 
IX.     Sines  and  Cosines,  Tangents  and  Cotangents        .        .        .  313 
X.     The  Tangent  and  Cotangent  in  the  Solution  of  Oblique- 
angled  Triangles 331 

XI.     General   Review  on  the   Solution   of  Triangles.     List  of 

Formulas  to  memorize      .         .         ....        .        .  340 

XII.     The  Quantity  V^I  in  Trigonometry 345 


CHAPTER   I.  , 

ELEMENTARY   DISCUSSION   OF   LOGARITHMS.  —  HOW   TO 
USE  TABLES  AND  WHAT  PLACE  TABLES  TO   USE. 

§  1.   Definition  of  a  Logarithm. 

The  expressions  ax  =  y,  (1) 

log#/  =  x,  (2) 

V  =  loga"1^  (3) 

indicate  the  same  relation  between  the  three  quantities,  a, 
x,  y;  namely,  that  x  is  the  logarithm  of  y.to  the  base  a.  The 
logarithm  of  any  number,  y,  to  the  base,  a,  is  the  exponent,  x, 
of  the  power  to  which  a  must  be  raised  to  produce  y. 

Expression  (2)  is  read,  "  the  logarithm  of  y  to  the  base  a 
is  x"  or,  as  is  usual  among  engineers,  "log  y,  base  a,  is  x." 
Expression  (3)  is  read,  uy  is  the  number  whose  logarithm  to 
the  base  a  is  x"  or,  more  briefly,  "y  is  anti-log  x,  base  a." 
Among  engineers  it  is  customary,  both  in  writing  and  in 
speech,  to  cut  the  word  logarithm  to  "log." 

Expression  (1)  is  called  the  exponential  form  for  a  loga- 
rithm ;   (2),  (3)  are  called  the  logarithmic  forms. 

23=  8,  (1),  is  equivalent  to  log28  =  3,  (2),  and  to  8  =  log2~13, 
(3);  and  102  =  100,  (1),  is  equivalent  to  log10100  =  2,  (2),  and 
to  100  =  log10_12,  (3).     Read  these  expressions. 

EXERCISES. 

1.  What   is   the   value   of   a10^?       Of   el0^?      Of    10lo«io*?      Of 

l0ga(l0ga-l*)  ?        Of    lOga"1  (}0gay)  ? 

2.  What  is  the  base  when  the  logarithm  of  81  is  4  ?  Express  this  in 
the  three  forms,  (1),  (2),  (3). 

3.  What  number  has  3  for  its  logarithm  when  the  base  is  6  ?  Express 
this  in  the  three  forms,  (1),  (2),  (3). 

1 


2  PLANE   TRIGONOMETRY.  [§  1 

4.  What  is  the  logarithm  of  1000  to  the  base  10  ?  Express  this  in 
the  three  forms,  (1),  (2),  (3). 

5.  Make  up  ten  examples  like  Exs.  2,  3,  4,  and  solve. 

6.  If  10  is  the  base,  what  are  the  logarithms  of  10,  100,  1000,  10000, 

10"  ?    If  0.01  =  —  =  —  =  10-2,  what  is  the  logarithm  of  0.01  ?     Of  0.1, 
100      102  »  & 

0.001,  0.0001,  0.00001  ?    Of  -A-? 

10» 

7.  Give  the  usual  illustration  from  algebra  from  which  is  drawn  the 
interpretation  that  a0  =  1.  In  accordance  with  this  interpretation,  what 
would  you  say  is  the  log  of  1  for  all  bases  ? 

8.  Express  in  the  form  of  an  identity  the  relation  of  a~x  to  ax.  How, 
then,  is  the  logarithm  of  a  number  related  to  that  of  its  reciprocal? 

9.  If  in  the  expression  ax  =  y,a  is  positive,  what  sign  has  y,  both  for 
positive  real  values  of  x  and  for  negative  real  values  of  x  ?  Give  some 
numerical  illustrations.  What  would  you  say,  then,  as  to  the  possibility 
of  negative  numbers  having  real  logs,  if  the  base  is  positive  ? 

10.  If  the  base  is  10,  between  what  limits  will  lie  the  logs  of  all 
numbers  between  1  and  10  ?  Between  10  and  100  ?  Between  100  and 
1000?  How  many  digits  to  the  left  of  the  decimal  point  have  numbers 
which  lie  between  10  and  100  ?  Between  1  and  10  ?  Between  100  and 
1000?  Write  three  four-figured  numbers  which  lie  between  1  and  10. 
Between  10  and  100.  Between  100  and  1000.  Calling  the  integral  part 
of  a  log  its  characteristic,  can  you  make  up,  from  your  answers  to  the 
preceding  questions,  the  rule  as  to  the  number  which  is  the  characteristic 
as  compared  with  the  number  of  digits  to  the  left  of  the  decimal  point 
in  the  number  whose  log  is  desired  ? 

11.  Express  0.1, 0.01, 0.001, 0.0001, 0.00001,  in  exponential  form  similar 
to  (10)-2  =  0.01,  where  the  base  is  10  and  the  exponents  negative.  From 
these  expressions  draw  conclusions  as  to  the  limits  between  which  lie 
the  logs  of  all  numbers  between  1  and  0.1,  and  between  0.1  and  0.01. 
Between  0.01  and  0.001.  Between  0.001  and  0.0001.  Write  four  num- 
bers which  lie  between  1  and  0.1.  Between  0.1  and  0.01.  Between 
0.01  and  0.001.  Between  0.001  and  0.0001.  What  sign  has  the  log  of  a 
decimal  fraction  ? 

It  is  customary  to  make  the  integral  part  of  such  logs  negative  and 
the  decimal  part  positive,  taking  for  the  integral  part  the  log  of  the 
power  of  10  next  below  the  given  decimal  whose  log  is  sought.  Can  you 
make  out,  from  your  answers  to  the  preceding  questions,  the  rule  as  to 
the  negative  number  which  is  the  characteristic  as  related  to  the  number 
of  zeros  to  the  right  of  the  decimal  point  before  a  figure  other  than  zero 
is  reached  in  the  number  whose  log  is  sought  ? 


§2]  DISCUSSION  OF   LOGARITHMS.  3 

§  2.   Working  Rules  for  Logarithms. 

(a)  The  logarithm  of  the  product  of  two  or  more  numbers  is 
the  sum  of  the  logarithms  of  the  numbers, 

or,  log  (mri)  =  log  m  +  log  n; 

log  (mnp)  =  log  m  +  log  n  +  log  p. 

Proof. 
If  ax  =  m,  (1),     then     logam  =  :z,  (2), 

and  if  av  =  n,    (3),     then     logan  =  y,  (4). 

Multiplying  (1)  by  (3),  ax+v  =  mn, 
or,  loga(mw)  =  x  +  y 

=  logam  +  logaw,  by  (2),  (4). 
Similarly  for  a  product  of  more  than  two  factors. 

EXERCISES. 

1.  If  n  is  a  positive  integer,  show  that  log  (an)  =  n  •  log  a. 

2.  Given  log102  =  0.30103  and  log103  =  0.47712,  find,  to  the  base  10, 
the  logs  of  the  following  numbers  :  6 ;  4 ;  9 ;  27 ;  8 ;  12 ;  36 ;  210 ;  38 ; 
2n ;  3m ;  2n  •  3m  (n,  m,  integers,  positive) . 

3.  Write  down  all  the  prime  numbers  from  1  to  100,  and  show  that  if 
the  logs  of  these  numbers  are  known,  the  logs  of  the  other  numbers 
below  100  can  be  calculated.  For  what  other  numbers  could  the  logs  be 
calculated  ? 

4.  If  e  is  the  base,  show  that  1  +  logen  =  \oge(en).  If  10  is  the  base, 
show  that  1  +  log10n  =  log10(10  •  n).     Show  also 

2  +  logen   =5  loge(e2  •  n)  ;  3  +  logen   =  loge(e3  -  ri)  ; 

n  +  logera   =  loge(en .  n)  ;  4  +  log10n  =  log10(  10000  •  ri) ; 

n  +  log10n  sa  log10(10"  ■  n). 

(5)  The  logarithm  of  an  indicated  quotient  [  —  )  is  the  loga- 
rithm of  the  numerator  minus  the  logarithm  of  the  denominator, 

or  log  ( — )  =  log  fn>  —  log  n. 


4  PLANE   TRIGONOMETRY.  [§2  6 

Proof. 
If  ax  =  m,  (1),     then     logam  =  x,  (2), 

and  if  ay  =  n,    (3),     then     logan  =  y,  (4). 

Dividing  (1)  by  (3), 

*—5    (5), 

or,  loga(^)  =  x-y 

=  \ogam  -  \ogan,  by  (2),  (4). 

EXERCISES. 

1.   If  log10  2  =  0.30103  and  log103  =  0.47712,  find  to  the  base  10,  the 

of  the  following  numbers:  |j  1.5;  4.5;  6.75;  1|;  2f;i;  ^;  f£; 

o  9     27     oTO 


(-)  (n,  m,  integers,  positive  or  negative). 


2.  Express  the  logs  of  the  proper  fractions  above  with  negative 
characteristic  and  positive  decimal  part  (mantissa)  to  the  log.  For 
example,  log  f  =  1.82331,  where  1  alone  is  negative  and  .82391  is  positive. 


.   Show  that  log  f  -  j  =  —  log  a. 


3 

4.   Show  that 


1  -  loge  x  =  logfl  -j  1  -  log10  n  =  log10  —  j 


x 


logex-  3  =  log.(^);  log10ft  -  5  =  log10(^). 

5.  Show  that  log10  27.34  =  log  2734  -  2  ; 

log]0  27.34  =  log  273.4-1; 
log10  3.415  =  log  3415  -  3. 

6.  In  general,  show  that  if  a  and  b  are  two  numbers  with  the  same 
sequence  of  digits,  but  with  the  decimal  point  differently  placed,  that 
a  =  10n  •  b,  and  that,  consequently,  the  logs  of  a  and  b  to  the  base  10  differ 
only  by  the  integer  n.  What  is  the  relation  between  the  decimal  parts 
of  the  logs  of  numbers  with  the  same  sequence  of  figures  ? 

7.  From  log10  2,  find  log105 ;  then  from  log10  3,  find  log1015. 

(<?)   The  logarithm  of  any  power  of  a  given  number  is  the 
logarithm  of  the  given  number  multiplied  by  the  exponent  of 
the  power, 
or,  log  (mp)  =p  log  m. 


3  a] 

DISCUSSION  OF  LOGARITHMS. 

Proof. 

If 

ax  =  m  (1),     \ogam=x,  (2) 

Then 

apx  =  mp,                                (3) 

or 

loga(mp)=px 

=  P  log«ro. 

Note.  —  We  assume  that  the  student's  knowledge  of  algebra  is  such 
that  he  will  grant  that  (3)  is  true  for  all  values  of  p,  positive,  nega- 
tive, integral,  fractional,  rational,  irrational.     Then, 

?       q 
log  (m«)  =  -  log  m,  (1) 

_?  a 

log(m  ')  =  -?.  log  ro,  (2) 

log  w«  =  log  Vm  =  -  log  m,        (3) 
n 

(1),  (2),  (3)  are  true  no  matter  what  base  is  taken.    (3)  is  so  important, 
it  may  be  set  in  words  as  a  rule : 

(<Z)  The  logarithm  of  the  nth  root  of  a  number  is  one  nth 
of  the  logarithm  of  the  number. 

§  3.    Special  Case  of  Logarithms  to  the  Base  10. 

(a)    Rule  for  Characteristic. 

105  =100000 

104  =10000 

103  =1000 

102  =100 

101  =10 

10°  =1 
10"1  =  0.1 

io-2  =  o.oi 
io-3  =  o.ooi 

10"4  =  0.0001 

io-5  =  o.ooooi 

The  logarithms  of  integral  powers  of  10  are  integers. 
Numbers  which  are  not  integral  powers  of  10  lie  between 


6  PLANE   TRIGONOMETRY.  [§3a 

consecutive  integral  powers  of  10.     Thus  the  logarithms  of 
such  numbers  lie  between  consecutive  integers. 
There  are  two  cases : 

(1)  Numbers  with  digits  to  the  left  of  the  decimal  point. 
Numbers  with  one  digit  to  the  left  of  the  decimal  point  lie 
between  1  and  10.  Their  logarithms,  as  the  table  on  page  5 
shows,  lie  between  0  and  1.  Numbers  with  two  digits  to 
the  left  of  the  decimal  point  lie  between  10  and  100.  Thus 
their  logarithms  lie  between  1  and  2 ;  that  is,  the  logarithm 
is  1  plus  a  decimal  fraction.  Similarly,  the  logarithm  of  a 
number  with  three  digits  to  the  left  of  the  decimal  point  is 
2  plus  a  decimal  fraction.  And  if  the  number  has  n  digits 
to  the  left  of  the  decimal  point,  its  logarithm  is  n  —  1  plus  a 
decimal  fraction. 

The  integral  part  of  a  logarithm  is  called  its  Characteristic  ; 
the  decimal  part,  its  Mantissa. 

For  numbers  with  digits  to  the  left  of  the  decimal,  the  char- 
acteristic is  the  positive  integer  one  less  than  the  number  of 
such  digits. 

The  characteristic  for  log  8  is  0 

The  characteristic  for  log  8.3         is  0 

The  characteristic  for  log  8.349     is  0 

The  characteristic  for  log  83  is  1 

The  characteristic  for  log  83.4       is  1 

The  characteristic  for  log  83.459  is  1 

The  characteristic  is  determined  wholly  by  the  number  of 
digits  to  the  left  of  the  decimal  point.  Those  to  the  right 
of  the  decimal  point  have  nothing  to  do  with  the  character- 
istic in  numbers  which  have  digits  to  the  left  of  the  decimal 
point. 

EXERCISES. 

The  teacher  may  assign  some  numbers  at  random,  and  question  as  to 
the  characteristic. 

(2)  Numbers  with  no  digits  to  the  left  of  the  decimal  point. 
All  such  numbers  which  have  no  zero  immediately  to  the 


§3  a]  DISCUSSION  OF   LOGARITHMS.  7 

right  of  the  decimal  point,  lie  between  1  and  0.1,  as  0.348, 
0.8925,  0.726952,  etc.  Thus  the  logarithms  of  all  such  num- 
bers lie  between  —  1  and  0.  Similarly,  all  such  numbers 
with  only  one  zero  immediately  to  the  right  of  the  decimal 
point,  as  0.043,  0.0987654,  etc.,  all  lie  between  0,1  and  0.01, 
and  have  their  logarithms  between  —  2  and  —  1.  And  so, 
in  general,  if  there  are  n  zeros  immediately  to  the  right  of 
the  decimal  point,  and  no  digits  to  the  left  of  the  decimal 
point,  the  logarithm  of  the  number  lies  between  —  (ti  +  1) 
and  —  n. 

The  characteristic  in  such  cases  is  usually  taken  as  the 
negative  number  which  is  the  larger  numerically,  so  that  the 
mantissa  for  such  logarithms  is  positive.  An  important  ex- 
ception is  mentioned  later. 

Thus,  to  get  the  characteristic  of  the  logarithm  of  a  number 
with  no  digits  to  the  left  of  the  decimal  point,  take  the  negative 
number  which  is  one  more  than  the  number  of  zeros  immediately 
to  the  right  of  the  decimal  point. 

The  sign  of  the  characteristic  is  set  immediately  over  it, 
when  negative. 

The  characteristic  of  log  0.23  is  1 
The  characteristic  of  log  0.2348  is  1 
The  characteristic  of  log  0.023  is  2 
The  characteristic  of  log  0.02345  is  2 
The  characteristic  of  log  0.00004  is  5 

EXERCISES. 

The  teacher  may  set  some  numbers  at  random  and  see  if  the  student 
can  state  the  characteristic  immediately. 

Note.  —  Instead  of  using  negative  characteristics,  many  calculators 
add  to  the  characteristic  such  a  multiple  of  10  as  will  make  it  posi- 
tive, and  allow  for  such  tens  in  the  calculation. 

For  1.7632  they  take  9.7632  -  10. 

For  2.7632  they  take  8.7632  -  10 ;  and  so  on. 


8  PLANE   TRIGONOMETRY.  [§3& 

This  always  seemed  to  me  an  unnecessary  procedure,  but  it  is  quite  com- 
monly followed  by  astronomers,  who  are  the  great  calculators.  It  is 
necessary  for  the  student  to  be  acquainted  with  both  processes,  as  some 
tables  and  some  books  follow  one  plan  and  some  the  other.  The  fact 
that  a  special  process  is  followed  by  the  astronomers  is  almost  a  proof 
that  it  is  best  as  a  calculation  process. 

(5)  Rule  for  the  Mantissa.  All  numbers  with  the  same 
sequence  of  digits  have  logarithms  with  the  same  mantissa. 
For  if  a,  5,  are  any  two  numbers  with  the  decimal  point 
differently  placed,  but  with  the  same  sequence  of  digits, 

a  =  6.10",  where  n  is  an  integer 

•'•  l°g10a  =  log106  +  rc. 

Thus  the  logarithms  of  a  and  b  differ  only  by  the  integer  n 
.-.  the  mantissas  are  the  same. 

For  this  reason  the  tables  give  only  the  mantissa  of  the 
logarithm  of  a  number.  The  calculator  supplies  the  proper 
characteristic. 

The  mantissa  is,  as  a  rule,  an  unending,  non-repeating 
decimal.  There  are  tables  which  give  the  mantissa  to  four 
places  of  decimals ;  to  five ;  to  six ;  to  seven  ;  to  ten.  A 
table  which  gives  the  mantissa  to  four  places  is  called  a  four- 
place  table;  and  so  on.* 

*  Tables  of  the  first  rank  are  : 

F.  G.  Gauss's  Five  Place  Tables  (German). 

Hussey's  Five  Place  Tables. 

Bremiker's  Six  Place  Tables  (German). 

Lodge's  Bremiker's  Six  Place  Tables  (English). 

Vega's  Seven  Place  Tables  (German). 

Vega's  Thesaurus  Logarithmorum,  or  Ten  Place  Tables  (published  1794 
and  still  considered  the  best  ten-place  table) . 

Zech's  Addition-Subtraction  Tables  (German). 

In  case  the  student  cannot  read  the  explanations  of  the  tables  which  are 
given  in  German  in  Gauss's  Tables,  Hussey's  Mathematical  Tables  will  be 
found  a  satisfactory  substitute.  In  my  judgment  no  five-place  table  surpasses 
that  of  F.  G.  Gauss.  This  table  has  been  made  the  basis  of  frequent  piracy, 
with  its  best  features  omitted.  His  Tables  cannot  be  changed  without  hurt- 
ing them.  This  is  the  general  opinion  of  all  calculators  who  have  used  them. 
For  this  reason  it  is  recommended  that  Gauss's  or  Hussey's  tables  be  used  in 
connection  with  this  book,  when  it  is  necessary  to  calculate  seconds  in  angles. 


§3c]  DISCUSSION  OF  LOGARITHMS.  9 

(<?)  Logarithms  of  roots,  when  the  characteristic  is  negative 
and  not  a  multiple  of  the  root  index. 

Given  log  0.002  =  3.30103,  what  is  log  V57002  ? 

(1)  First  Method  (characteristic  negative). 

log  V0^02  =  1  log  0.002  =  3-30103  =  i  +  1-80108  =  2.65052 

A  '- 

The  general  process  is  to  bring  the  characteristic  to  the 
next  integer  above  itself  into  which  the  root  index  will  go 
exactly,  balancing  this  subtraction  by  the  addition  of  the 
corresponding  number  to  the  positive  mantissa,  as  in 

log  ^/<U)002  =  i-30103  =  6±2^0103  =  ^mi 
o  o 

4.30103 


log  V0.0002  =  ^-°^w  =  1.47158 

(When  the  sixth  figure  of  decimals  in  the  quotient  is  5  or 
more,  increase  the  terminal  figure  (fifth)  by  unity.) 

EXERCISES. 

The  teacher  may  assign  them  at  will. 

(2)  Second  Method  (characteristic  positive  by  aid  of  tens). 


log  V0.002 

l.dvivd  — 

1U 

2 
17.30103 

-20 

2 

=  8.65052- 

6.30103- 

3 
26.30103 

10 
10 

-30 

log  \/0.0002 

3 

=  8.76701- 
6.30103- 

7 
66.30103 

■10 
•10 

log  -n/0. 0002 

-70 

7 
=  9.47158- 

■10 

10  PLANE   TRIGONOMETRY.  [§3tf 

In  general,  a  multiple  of  ten  one  less  than  the  root-index 
is  added  and  subtracted. 

EXERCISES. 

The  teacher  may  assign  some  at  will,  selecting  also  some  fractional 
indices  which  modify  the  general  rules  above. 

(cT)  Logarithms  when  negative  factors  occur. 

Since  a~x=  — » 

ax 

in  the  expression  a±x  =  y, 

if  a  is  positive,  y  is  positive  for  negative  real  values  of  x  as 
well  as  for  positive  real  values  of  x. 

Thus,  with  a  positive  base  negative  numbers  have  no  real 
logarithms. 

Logarithms  can  be  used,  however,  in  getting  products,  quo- 
tients, powers,  roots,  of  negative  numbers,  by  using  the  loga- 
rithms as  if  the  numbers  were  positive,  attaching  the  proper 
sign  to  the  result. 

It  is  customary  to  write  (n)  after  the  log  of  the  factor 
which  is  negative.  An  even  number  of  n's  will  indicate  a 
positive  result  and  an  odd  number  a  negative  result. 

EXERCISES. 

1.  Find  the  value  of  a'b'c,  where  a  =  17,  b  =  13,  c  =  —14,  d=  -  23, 
c=-9.  d'e 

The  result  is  negative.    The  process  is  : 

log  17  = 

log  13  =  log  23  =                    (n) 

log  14  = (n)  log    9  = (w) 

Sum=                    (1)  Sum=                    (2) 

Subtract  (2)  from  (1)  and  look  up  in  the  tables  the  number  correspond- 
ing to  the  resulting  logarithm.  (Directions  for  doing  this  will  be  found 
in  the  accompanying  tables.) 

2.  Given  log10  2  =  0.30103,  and  log10  3  =  0.47712,  find  logs  of  following 
numbers  to  base  10 :    28,    2§,    2*,    2*,    3?,    35,    3s,    V%,    \/02,    v'oAM, 

y/mm,   tyumm,  (-12)7,   (-6)1,   (-2)*,   (-2)-t,  (-0.02)!, 

(_2)3(_3)8,  (2)*(-3)'. 


§3e]  DISCUSSION   OF  LOGARITHMS.  11 

(e)  Use  of  the  Cologarithm  in  Calculations.  —  The  loga- 
rithm of  the  reciprocal  of  a  number  taken  so  that  the 
mantissa  is  positive  is  called  the  cologarithm  of  the  number. 
The  characteristic  may  be  positive  or  negative.  When 
negative  it  can  be  made  positive  by  the  addition  and  sub- 
traction of  an  appropriate  multiple  of  10,  if  one  is  following 
the  calculation  process  pointed  out  in  the  Note  on  page  7. 
In  such  procedure  one  must  allow,  in  the  final  result  of  a 
summation  of  cologarithms,  for  the  tens  thus  introduced. 

colog  (a)  =  logQ  =  -  log  (a). 

log  20  =  1.30103  ;  colog  20  =  2.69897,  or  8.69897  -  10. 
log  0.02  =  2.30103  ;  colog  0.02  =  1.69897. 

Rule  for  writing  the  cologarithm  of  a  number:  Subtract 
each  digit  of  the  mantissa  from  9,  from  left  to  right,  except  the 
terminal  digit  to  the  right.  Subtract  this  from  10.  Add  1, 
algebraically,  to  the  characteristic  and  change  the  sign. 

In  calculating  the  value  of  — - — — ,  one  may  add  the  logs 

oi  *  e 

of  a,  5,  and  c.  Also  add  in  a  separate  column  the  logs  of  d,  e. 
Then  subtract  the  second  sum  from  the  first,  and  look  up 
the  number  corresponding  to  this  log.  Or  by  the  aid  of 
cologarithms,  the  calculation  may  all  be  carried  out  in  one 
column,  as  follows : 

log«  = 

log  5  = 

log<?  = 
colog  d  = 
colog  e  = 

Sum  = 

EXERCISES. 

1.  Given  log  2  =  0.30103,  find  cologs  of  20 ;  800 ;  4,000,000. 

2.  Given  log  3  =  0.47712,  find  cologs  of  0.3 ;  0.003 ;  K/Q&. 

Whether  to  use  the  colog  or  not  depends  on  circumstances. 
Frequently  it  is  best  not  to  use  it ;  frequently  it  lends  itself 


12  PLANE   TRIGONOMETRY.  [§3e 

to  a  handy  calculation-scheme.     For  example,  later  we  shall 
use  the  expression, 

a     b      c 
where  x,  y,  z  are  the  sines  of  the  angles  of  a  triangle.     If  5,  c 
have  to  be  calculated, 


b  =  y--;            c=z 

:  ._. 

X 

X 

And  the  calculation-scheme  is, 

log  5  = 

(6) 

\ogy  = 

(1) 

logtf  = 

(2) 

colog  X  = 

(3) 

log  25  = 

(4) 

logc  = 

(5) 

where  (6)  is 

(l)  +  (2)  +  (3) 

and  (5)  is 

(2)  +  (3)  +  (4> 

I 

one  and  the  same 

scheme  giving  5,  c. 

In  adding,  cover  the 

log  not  to  be  used 

(/)  Using  tables  of  logarithms. 

It  is  recommended  that  at  this  point  the  student  be  taught  how  to  look 
up  the  logarithms  of  numbers  in  a  four-place  table  and  in  some  five-place 
table,  both  in  finding  the  logarithms  corresponding  to  numbers  and  in 
finding  the  numbers  corresponding  to  logarithms,  taking,  for  the  present, 
only  the  cases  where  the  logarithms  and  numbers  can  be  found  exactly, 
or  to  a  close  approximation,  in  the  tables,  without  interpolation.  The 
principles  on  which  interpolation  are  based  can  be  understood  only  after 
further  study,  as  set  forth  in  the  next  several  articles. 

The  student  should  carry  out  the  operations  with  the  tables  and  with- 
out the  tables  and  compare  results.  How  to  use  the  four-place  tables  of 
this  book  is  explained  in  the  preface  to  the  tables. 

EXERCISES. 

(For  a  four-place  table.) 
Calculate  with  logarithms  and  without  logarithms  the  values  of  the 
following  expressions  and  compare  results.     Give  results  to  not  more 
than  three  figures  and  to  only  the  nearest  approximation  which  the 


3  4]  DISCUSSION  OF  LOGARITHMS.  13 

tables  will  give  without  interpolation.  In  the  case  of  fractions,  calculate 
their  values  with  cologs  and  without  cologs,  using  negative  character- 
istics and  also  characteristics  positive  by  the  aid  of  10's.  Find  b,  c, 
in  Exs.  4,  5,  6. 

1.  2x3;    3x4;    4x5;    5x6;    6x7;    7x8;    8x9;    9x10; 
10  x  11 ;    11  x  12 ;     12  x  13. 

2.  2x3x4;  3x4x5;  4x5x6;  5x6x7;  6x7x8;  7x8x9; 
8x9xl0;2x3x9;3x4x9;4x5x9;5x6x8. 

3    12  x  13.  13  x  14.  16  x  17.  18  x  19.  25  x  27.  35  x  37.  45  x  47 
7       '        9       '       11      '       13      '       32      '       43      '       53 

,    45  x  49  x  85  .    56  x  57  x  73  .    96  x  83  x  85      0.21     0.41     0.51 


5. 


7. 


71  x  73      '        81  x  91      '        61  x  77      '    0-31        b    "    c 
243  x  244  .    256  x  295  .    259  x  831 .    0.031  =  0.023  =  0.033 
231       '         331       '         923       '    0.016         b  c 

341  x  441  x  551 .    349  x  837  x  624  .    .317  =  .231  _  .314 
536  x  437       '         555  x  989       '   .218        b  c 

VMl  •  #225       #0034  •  VO00l4 


#316  •  #719     #0^015  •  #0.00042 

(_95).  (-34).  (-65),  (-32).  (-41).  (-51).  (64) 
27- (-86)  '        (-17).  (-29).  (-43) 

9.   Find  to  three  figures : 

#2;  #3;  #5;  #6;  #2;  #3;  #4;  #5;  #2;   #3;  #5;  #6. 

(6.6)8.  (1.02)* .  (2.51)6 .  (3.03)8 
(2.5)5  •  (3.15)2 '  (2.47)8  •  (1.78)4* 


8 


10. 


J(3.11)«.(2.71)'.    '/(47)».(-85)« 
'    ^(2.14)5.(3.19)8'    *    (23)2.  (71)4  ' 

§  4.    Systems  of  Logarithms. 

While  any  finite  positive  number,  other  than  1,  may  be 
made  the  base,  there  are  only  two  bases  of  importance. 
These  are  10  and  the  number  indicated  by  the  letter  e,  where 

e  =  2.7182818284-... 

The  decimal  part  of  e  is  unending  and  non-repeating. 
The  number  e  is  the  limit  toward  which  the  expression 


H)' 


14  PLANE   TRIGONOMETRY.  [§  4 

approaches  as  x  becomes  infinite  ;  that  is,  as  will  be  shown 
presently,  e  is  the  limit  toward  which  the  infinite  series 

approaches,  as  the  number  of  terms  approaches  infinity. 
The  logarithms  with  the  base  10  are  called  the  Briggs  Sys- 
tem, or  Common  Logarithms.  The  logarithms  with  the  base 
e  are  called,  in  honor  of  Baron  Napier,  the  inventor  of 
logarithms,  the  Napierian  System.  They  are  also  called  the 
Natural  Logarithms.* 

§  5.    Changing  from  One  System  to  Another. 

Logarithms  are  first  calculated,  in  the  manner  shown  in 
§  10,  to  the  base  e.  From  the  table  to  the  base  e,  the  table 
to  the  base  10  is  then  calculated,  by  dividing  each  log  in  the 
e-table  by  the  log  of  10  in  the  e-table.     That  is,  the  number 

— -   used  as  a   multiplier  for  the  e-table  will  give  the 

loge10 

10-table. 

The     number      •    *      =  0.4342944819    •  •  .  is    called   the 
loge10 

Modulus  of  the  Briggs  System,  or  Common  Logarithms,  with 

reference  to  the  Napierian  Logarithms. 

In  general,  any  table  of  logs  can  be  converted  into  a  table 

to  another  base  by  dividing  every  log  in  the  table  by  the 

log  of  the  new  base  in  the  given  table. 

Proof. 

Let  ax  =  m  =  by,  (1) 

so  that  x  =  \ogam,  (2) 

and  y  =  \0gbm.  (3) 

*  As  indices  had  not  found  a  place  in  Algebra  at  the  time  of  Napier,  his 
logarithms  were  very  different  from  those  now  called,  in  his  honor  merely, 
Napierian  Logarithms.  The  student  interested  in  the  subject  may  consult 
Cajori's  "History  of  Mathematics"  ;  the  section  "Logarithms"  in  the  Ency- 
clopaedia Britannica ;  also  article  by  J.  W.  Young  in  Am.  Math.  Monthly,  1903. 


§6]  DISCUSSION   OF   LOGARITHMS.  15 

Take  the  logarithm  of  (1)  to  the  base  a, 

.-.  x  =  y\ogabi  (4J) 

or  y  =  -?—.  (5) 

By  (5),  logarithms  of  the  5-system  are  those  of  the 
a-system  divided  by  the  logarithm  of  b  in  the  a-system  ; 
whence  the  general  rule  for  changing  logarithms  of  a  given 
system  to  a  new  system  : 

Divide  the  logarithms  of  the  given  system  by  the  logarithm 
of  the  base  of  the  new  system  taken  from  the  old  system. 

The  number, -,  is  called  tjie  modulus  of  the  6-system 

loga& 

with  reference  to  the  a-system. 

Thus  the  modulus  of  the  Common  Logarithms  with  ref- 
erence to  Napierian  Logarithms  is 

-JL-  =  0.43429  •  •  -. 
log,  10 

And  the  modulus  of  the  Napierian  Logarithms  with  refer- 
ence to  Common  Logarithms  is 

— —  =  2.303.... 

logio  e 

Tlius,  to  convert  a  Napierian  table  to  a  Common  table,  multiply 
each  logarithm  by  0.43429 

And  to  convert  a  Common  table  to  a  Napierian  table,  mul- 
tiply each  logarithm  by  2.303 

§  6.    loga  b  •  log6  a  =  1. 
Proof. 
Let  ax=b,  (1) 

so  that  loga  b  —  x.  (2) 

Take  the  logarithm  of  (1)  to  the  base  5, 

.-.  x  •  log6  a  =  1,  (3) 

.-.  by  (2),  loga6.1og6a=l,  (4) 


16  PLANE   TRIGONOMETRY.  [§  6 

Thus  the  modulus  of  the  5-system  with  reference  to  the 
a-system  is  either -,  or  log6a. 

In  particular, — -  =  log10  e. 

log,  10 

Thus  the  modulus  of  the  Common  system  is  either  the 
reciprocal  of  .the  logarithm  of  10  in  the  e-system,  or  it  is  the 
logarithm  of  e  in  the  10-system. 

EXERCISES. 

1.  Look  up  loge  in  tables  and  compare  with  the  modulus. 

2.  How  would  you  convert  an  ordinary  table  of  logs  to  the  base  10 
into  a  Napierian  table  ? 

Ans.    Divide  each  log  by  0.4342  •  •  •,  or  multiply  each  by  2.303  •  •  •. 

3.  Of  what  number  in  the  10-system  is  the  modulus,  0.4342  •  •  •,  the 
log  ?    Of  what  number  in  the  e-system  is  2.303  •  •  •  the  log  ? 

ax  =  m  =  by, 
.  log5  m  •  logc  b 

&  logc  a 

log10  37.31  =  1.57183 
log10  2  =  0.30103, 
log2  37.31  and  log2  373.1 
log2  3731  and  log2  37310. 

logj0  e  =  0.43429, 
log10  2  =  0.30103, 
log10  3  =  0.47712, 
log10  7  =  0.84510, 
find  to  the  base  e,  the  logs  of  the  following  numbers :  2 ;  3 ;  4 ;  5 ;  8 ; 

9;  is  f ;  I;  I;  2f;  xiu- 

7.  By  the  aid  of  the  preceding  logarithms,  find  x  in  the  expressions : 
2*  =  3;  3*  =  2;  5*  =  12;  16*  =  10;  27*  =  4. 

(a)  Using  e-logs.     (b)  Using  10-logs. 

8.  Given  log,    2  =  0.69315, 

log6    3  =  1.09861, 
and  loge  10  =  2.30259, 

find  to  base  e  the  logs  of  following  numbers :  6 ;  12 ;  4 ;  9 ;  8 ;  27 ;  5 ;  15 ; 
25;  125;  f ;  f;  |;  f. 


4. 

From  the  relation 

show  that 

5. 

If 

and 

find 

1 

and 

6. 

Given 

§7] 


DISCUSSION  OF   LOGARITHMS. 


17 


9.   From  the  values  of  loge  2,  loge  3,  given  in  Ex.  8,  calculate  log10  2, 
log10  3,  and  compare  with  values  previously  given. 

10.   If  ex  =  V#2+  1  +  y,  then  e*  -  e~*  =  2  y ;  ex  +  e~x  =  2Vy2  +  1. 


2y;  e*  +  e-* 
I,  then  ex  +  e~z  =  2  y ;  e*  -  c~x  =  2Vy2  -  1. 


11.   If  e«  =  y  -  vy 
§  7.    The  Base  of  the  Napierian  System  of  Logarithms. 
e  =  limz=Jl  +  -J  • 
By  the  binomial  theorem,  when  y  <  1, 


(l+y)--l+*y  +  *l?LrJ 


V  +  etc., 


1-2-3 


■■■  {}  +  -)  ^^-y +-TTW  +    1-2-3    0  +  etc- 

_i .  1  +  1 *J ,  \    *A    */  ,  etc 

-1  +  i+     1.2     +         1-2-3         +  etc- 
Let  x  approach  the  value  infinity, 

liuw(l  +  iy=l  +  l+l  +  l  +  l+etc. 

The  value  of  e  may  be  calculated  to  five  places  as  follows : 

1.000000 


1.000000 


0.500000 


0.166667 


0.041667 


0.008333 


0.001388 


0.000198 


0.000025 


0.000003 


e  =  Sum  =  2.71828 

To  ten  places,  e  =  2.7182818284  .... 

e  is  an  irrational  number,  or  an  unending,  non-repeating 
decimal. 


18 


PLANE   TRIGONOMETRY. 
COROLLARIES. 


[§7 


1. 

2. 
3. 
4. 
5. 
6. 
7. 
8. 


x/x=<x>     e 


9.  ft- AY  =±. 

l-iy*=fl  +  -T=«.         10.    (1  +  xY     =e. 

IV    Bi( 

1\na; 

-)         =6n. 

Xjx=«> 

x)x=*>     en 


1  + 


i-*Y     = 


xj 


l+iy  =^. 

nx/x*=*> 


11. 

(1+*)  *=-. 

12. 

(1  +  nxY    =en 

n 

13. 

(l  +  x)l=e*. 

14. 

(1  —  nxf     =  — 

V                   yx=0        en 

15. 

\            tf/a;=0 

16. 

ri-;#  =  A_ 

-w 


§  8.     The  Exponential  Series,  or  Exponential  x. 

e*=l  +  y  +  j|  +  j|  +  ^+etc., 

0  =  lim^Jl  +  -J  , 

.-.     ^  =  limx=^l+^ 
or,  by  the  binomial  theorem, 

e»  =  lim_  of  (l  +  ^g)+^Mlll)gJ+  etc.) 

-S  y-('-i)(y-P 


=  limr_^  of 


l  +  y  + 


yy 


1-2 


1.2-3 


+  etc. 


§8]  DISCUSSION   OF   LOGARITHMS.  19 

V2       V3       V4 

or,  *y=1+^  +  !+J3+|I+  etc* 

x2      x3      at 
Similarly,  ex=l  +  x  +  —  +.-4-.-T  4-  etc. 

E    12    Iz  -  % 

This  series  is  called  the  Exponential  Series,  or  Exponen- 
tial x.  It  is  true  for  all  finite  values  of  x,  that  is,  it  converges 
to  a  definite  limit  for  all  values  of  a?,  except  #=oo .  No  matter 
how  large  x  is,  if  finite,  the  numbers  in  the  factorial  denomi- 
nators finally  overtake  it  in  value.  From  that  point  on,  the 
terms  decrease  in  value.    Thus,  the  ratio  of  the  general  term, 

—   to  the  preceding  term,  — — -,  is  , — h: r  =  --      This 

\n  \n  —  l         [w      \n  —  1      n 

ratio  has  the  limit  zero,  when  n  =  oo,  for  any  finite  x.     Thus, 
the  series  is  convergent.     (See  any  College  Algebra.) 

Note.  —  The  teacher  may  find  it  advisable  here  to  give  the  proof  of 
the  convergency  of  an  infinite  series,  as  covered  by  the  statements : 

(a)  If  the  ratio  of  the  nth  term  to  the  preceding  is  <  1,  when 
n  =  co  ,  the  series  is  convergent. 

(b)  If  this  ratio  is  >  1,  the  series  is  divergent. 

(c)  If  this  ratio  is  =1,  the  matter  is  in  doubt. 


EXERCISES. 

1. 

1?    l?  +  li    15   15 

2. 

^  +  e~l)  =  1  +  tttteUi' 

3. 

^ttt^^ttt^1' 

4. 

H+H+^1+(1+H+^)* 

5. 

l  +  2  +  3      A  =  e.                      6>    2       4  +  6                 1 

15    15    II    2                   IS    15    II           e 

7. 

ttt^     e-l 

1+ttet°-     e  +  1 

20  PLANE   TRIGONOMETRY.  [§  9 

8.    1 +^3  +  §!  +  ^  +  etc.  =  5e. 


IS    IS    12 

+21+2+31+2+3+4.     _3 
12     +        [8         +  |4  +^-"2 


§  9.     The  Logarithmic  Series. 

loge(l  +  z)=z--+----- 
8eV        /  2      3      4 


-ioge(i-s)=s+!+!+j 

Proof. 

If 

^=1+3, 

then 

iog«(i+*)=y- 

But 

.-(i4Y  • 

\             X/X=<x> 

And 

\             XJx=co          \              XJx=<*> 

•••   by  (1), 

-         1/            .              1 

(1) 


ey. 


.-.  1  +  -  =  (1  +  g)«,  when  x  =  cc. 

X 

1 

If    55  <  1,   we    may    expand    (1  +  z)*    by    the    binomial 
theorem. 

1/1      .A         1/1 

•-•1  +  ;-I+;*+iT72 


Cancel  the  l's  and  multiply  by  x. 

,e-'>,q--)(H,lrt: 

•■•*"J+      1.2      +        1.2-3  +         ' 

when  #  =  oc. 

Z2    ,    Z3       z4       . 

•*^==2-2+3"4'etC-' 
or  loge  (1  +  «)  =  i  -  |  +  |3  -  J,  etc.  (1) 


§  10J  DISCUSSION   OF  LOGARITHMS.  21 

This  series  has  been  derived  under  the  supposition  that 
z  <  1,  numerically,  and  is  therefore  subject  to  that  limitation. 
Changing  z  to  —  s, 

loge  (1  -  0  =  -  (z  + 12  + 13  +  J  +  etc.).  (2) 


§  10.    Calculation  of  Napierian  Logarithms. 

The  series  (1),  (2),  above,  are  not  adapted  to  the  calcula- 
tion of  logarithms.  They  hold  only  when  z  is  numerically 
less  than  1  ;  (1)  holding  also  when  z  is  1.  And  even  within 
the  field  of  their  convergency,  they  are  very  slowly  conver- 
gent, that  is,  a  large  number  of  terms  have  to  be  taken  to 
get  an  approximate  value  of  the  logarithm  of  a  proper  fraction. 

A  rapidly  convergent  series,  suitable  for  computation  pur- 
poses, is  obtained  thus  : 

Subtract  (2)  from  (1). 

•••io^:=2[z+f+f+7+etc-)    (3) 

Since  z  is  to  be  less  than  1,  let 

1 

where  a:  is  a  positive  integer. 

_  1  +  2        +  2z  +  l      2z  +  2     s  +  1 

Inen  q = = —  =  — ~ = 

1—2      -  1  zx  x 

2x  +  l 

Now  x  and  x  +  1  are  consecutive  numbers,  when  x  is  an 
integer. 

And      log  £±1  =  log  (x  + 1)  -  log  x  =  log  1±£ 

•'.  log,  (x  +  1) 


22  PLANE   TRIGONOMETRY.  [§  10 

As  soon  as  x  is  large,  a  very  few  terms  of  this  series  will 
give  loge  (1  +  x)  to  a  close  approximation,  when  loge  x  is 
known. 

Since  loge  1  =  0,  the  Napierian  table  is  constructed  as 
follows : 

loge3  =  lo&2  +  2g  +  i.l  +  ^.|6  +  etc.) 

log,  4  =  2  loge  2, 

log,5  =  log,4  +  2g  +  l.|  +  l.|+etc). 
And  so  on. 

The  number  of  terms  taken  in  the  series  is  dependent  upon 
the  place  of  the  table  to  be  constructed. 

EXERCISES. 

1.  Calculate  to  four  decimals  loge2;  loge  3;  loge4;  loge5;  loge6; 
loge  7;  loge  8;  loge  9;  loge  10.  Reduce  each  such  logarithm  to  the  base 
10,  and  compare  with  the  logarithms  in  the  table  accompanying  this  book. 


f  loge  2  =  0.6951 

Ans.<  loge 3  =  1-0986 

[loge  4  =  1.3863 


loge  5  =  1.6094 
loge  6  =  1.7918 
loge  7  =  1-9459 


loge  8  =  2.0794; 
loge  9  =  2.1972 ; 
loge  10  =  2.3026. 


2.  What  relation  has  loge  10  to  the  modulus  of  the  e-system  as  com- 
pared with  the  10-system?  What  relation  has  it  to  the  modulus,  0.4343, 
of  the  Common  system  as  related  to  the  e-system  ? 

§  11.    Series  for  the  Logarithm  of  Any  Number  whatever,  without 
Reference  to  the  Adjacent  Number. 

If  in     +  g  we  let  z  =  x~~  .,  so  that  z  <  1,  if  x  is  any  posi- 
1—2  x+1 

1  +  — 

..  ,  1  +  2  X  +  1 

tive  number,  q = T  =  x. 

1  —  2        -j         X—l 

x  +  1 


§  12]  DISCUSSION  OF  LOGARITHMS.  23 

.-.  Series  (3)  becomes 

,1o8„«.o.8«(|^+|(|^)vi(Si;+.,4W 

Series  (5),  (6)  define  what  is  meant  by  logex  and  log10x 
as  related  to  ar,  in  calculation  form,  0.8686  being  twice  the 
modulus,  approximately. 

EXERCISES. 

1.  Show  that  1  -  $  +  $  -  i  +  i-"  =  loge2.  Is  this  the  best  series 
from  which  to  calculate  loge2?    Is  it  rapidly  or  slowly  convergent? 

2.  Show  log.8-lo«j.2=i-|.i  +  |.i-|.l,  etc. 

3.  Show  that  if  x  >  1,  series  (3)  becomes 

4.  Show  that  log, a  -  loge ft  =  «^i  + 1  f^zl) *  +  1  («Z^  8+  etc. 

a         2  \    a    I       3  \    a     / 

5.  Show  that  loge  (1  +  3* +  2z2)  =  3x  -  ^+ ^  -  i^+ ... 

^  O  TC 

if  2  a:  be  not  >  1. 

6.  Show  that    2  log,  a:  -  log.  (x  +  1)  -  log,  (x  -  1) 

7.  Show  10^2=^  +  ^ +  ^+.. .. 

8.  Show  log.2  - 1  =  _!_  +  _1_  +  _|_  +  .... 

§  12.    The  Rule  of  Proportional  Parts  in  Logarithms. 

Let  x  and  x  +  e  be  two  numbers,  where  e  is  small  compared 
with  x. 

Then  loge  (x  +  e)  -  log,  rr  =  loge  ?^p  =  loge  f  1  +  ^J . 
Butloge(l  +  i)  =  i-lgJ+lg)3,etc.(§9). 


24  PLANE  TRIGONOMETRY.  [§  12 

If  -  is  small,  (-) »  (~m  are  s^  smaller.     For  instance,  if 

-  =  o.ooooi,  f-)2=  0.0000000001. 

x  \xj 

.*.   loge  ( 1  +  -  ]  =  - ,  approximately, 

and  log10  (l  +  j)«  0.4343  U*  -e,  nearly, 

where  iJf  is  the  modulus. 

x     i  0.4343 

.'.   log10(s  4-  e)  -  log10  a;  -  — —  e,  nearly. 

TAws  £A#  difference  of  the  logs  of  any  two  numbers  which 
differ  by  a  quantity  relatively  small,  is  proportional,  approxi- 
mately, to  the  difference,  e,  of  the  numbers. 

This  is  called  the  Rule  of  Proportional  Parts. 

In  the  tables  of  logarithms  the  numbers  differ  by  unity. 
Thus  e  =  1,  and 

M    0.43429...  .    .,     -.  .      _. 

—  = is  the  Tabular  Difference, 

xx  M 

in  tables  to  the  base  10. 

Thus  the  tabular  difference  is  large  at  the  beginning  of 
the  table,  and  decreases  as  the  numbers  increase.  Also,  a 
given  difference  in  numbers  at  the  beginning  of  the  tables 
will  make  a  larger  difference  in  the  logarithms  of  the 
corresponding  numbers  than  will  the  same  difference  in 
numbers  occurring  at  any  other  part  of  the  table. 

Four-place  tables  are  generally  arranged  to  give  directly 
the  logarithms  of  three-figured  numbers,  and  five-place 
tables  (arranged  like  Gauss's)  give  directly  the  logarithms 
of  four-figured  numbers.  Since  the  differences  in  logarithms 
at  the  beginning  of  the  table  are  large  and  vary  rapidly,  it 
is  advisable  that  a  four-place  table  should  give  directly  the 
logarithms  of  four-figured  numbers  at  the  beginning  of  the 
table.  The  table  in  this  book  follows  this  plan  in  numbers 
up  to  1709. 


§  13  a]  DISCUSSION   OF  LOGARITHMS.  25 

§  13.    Applications  of  Tabular  Difference. 

M 
By  §  12,  loge(>  -f-  e)  -  loge  a;  =  —  e, 

x 

or  log-difference  =  tabular  difference  times  number-difference. 

(a)   To  find  the  logarithm  of  a  number  not  in  the  table. 
Given         log  2054  =  3.31260 
and  log  2055  =  3.31281.  What  is  log  2054.2  ? 

The  tabular  difference  is  21. 
The  number-difference  is  0.2. 
.*.  the  log-difference  is  0.2  of  21,  or  4.2,  or  4. 
.-.  log  2054.2  =  log  2054  plus  0.00004  =  3.31264. 

Note.  —  The  usual  rule  for  "  throwing  away  "  is  followed  in  the  log- 
differences.  The  above  difference,  4.2,  is  counted  as  4.  If  it  had  been 
4.8  or  4.5,  it  would  have  been  taken  as  5.  When  the  number  to  the 
right  of  the  point  is  less  than  5,  it  is  thrown  away ;  when  it  is  5  or  more 
than  5,  the  number  to  the  left  of  the  point  is  increased  by  1. 

Since  the  mantissa  with  the  base  10  is  independent  of  the 
position  of  the  decimal  point  in  the  number,  when  the  loga- 
rithm of  a  number  is  to  be  interpolated  from  the  table- 
logarithms  a  decimal  point  is  set  in  the  given  number  after 
the  number  of  digits  for  which  the  table  gives  the  logarithm. 
For  instance,  if  the  table  gives  logarithms  of  numbers  with 
four  digits,  and  the  logarithm  of  2.3459  is  desired,  consider 
this  number,  so  far  as  mantissa  is  concerned,  as  2345.9,  and 
interpolate  for  0.9,  as  above  for  0.2.  Similarly,  for  log 
456789  interpolate  for  0.89  from  log  4567  and  4568. 

Generally  an  interpolation  for  more  than  two  figures  can- 
not be  made,  since  if  numbers  differ  by  0.001  or  by  0.009 
beyond  the  table  figures,  their  logarithms  agree  to  the  place 
of  the  table.  Often  the  second  figure  is  a  matter  of  indif- 
ference. 

4343 

For,  log10  (x  +  e)  -  log10  x  = : e  ; 

x 

4343 
or,  log-difference  =  : times  number-difference,  and  num- 

x 

ber-difference  =  2.3  x  times  log-difference. 


26  PLANE   TRIGONOMETRY.  [§13  a 

Thus  the  smaller  x  is,  the  larger  is  the  log-difference  for  a 
given  number-difference.  If  we  are  using  a  five-place  table, 
like  Gauss's,  giving  logarithms  of  four-figured  numbers,  the 
smallest  value  of  x,  so  far  as  mantissa  is  concerned,  is  1000, 
and  in  a  five-place  table  the  smallest  appreciable  log-differ- 
ence is  0.000005.  Thus  if  the  number-difference  were  0.009 
in  the  neighborhood  of  1000,  the  log-difference  would  be 
0.0004342  times  0.009,  or  about  0.0000039,  not  enough  to 
affect  the  logarithm  in  the  fifth  place. 

Thus  in  no  table  (which  gives  directly  logarithms  for 
numbers  of  one  figure  less  than  the  place  of  the  table)  is  it 
possible  to  interpolate  for  more  than  two  figures  beyond  the 
reading  of  the  table  ;  often  not  for  a  second  figure. 

(b)  To  find  the  number  corresponding  to  a  logarithm  lying 
between  two  logarithms  in  the  table. 

Given  3. 74687  =  log10  5583 

and  3.74695  =  log10  5584, 

1.74689  =  log10? 

The  tabular  difference  for  a  unit  number-difference  is  8. 
The  log-difference  between  the  given  log  and  that  next  under 
it  in  the  table  is  2. 

__      ,        _.„  log-difference 

Number-difference  = - — 

tabular  difference 

=  f  =  i  =  .25. 

.-.  1.74689  =  log10  55.8325. 

The  characteristic  fixes  the  position  of  decimal  point. 

To  how  many  places  should  such  divisions  be  carried  ? 

This  depends  upon  the  number  x. 

For,  by  the  equations  above, 

e  __  log-difference 

x~~    0.43429-.. 

For  a  five-place  table  the  smallest  appreciable  log-difference 

is  0.000005 ;  that  is,  5  in  the  sixth  place. 

•"• here  I = raif  -  » t0  le  appreeiabh- 


§13  6]  DISCUSSION   OF   LOGARITHMS.  27 

Thus  in  using  a  five-place  table,  the  division  should  not  be 
carried  to  a  place  where  1  in  that  place  bears  to  the  number 
being  found  a  ratio  smaller  than  1  to  86859  (e  to  x). 

For  instance,  above  we  carried  the  division  to  two  places, 
getting  25,  where  the  number  found  was  55.8825.  The  ratio 
of  1  in  the  final  place  to  the  number  is  1  to  558325.  This  is 
less  than  1  to  86859.  Thus  the  division  was  carried  too  far. 
In  fact,  since 

5    <  x 


558325      86859' 

the  final  5  would  not  affect  the  logarithms  to  five  places. 
That  is,  the  logarithm  of  558325  is  the  same  as  that  of  55832 
to  five  places. 

The  general  rule  for  any  table  giving  logarithms  directly 
for  numbers  of  one  figure  less  than  the  table  is : 

Do  not  carry  the  division  to  the  place  where  1  in  that  place 
has  to  the  number  being  found  a  ratio  less  than  1  to  the  double 
modulus,  considered  as  an  integer  to  as  many  figures  as  the 
place  of  the  table. 

Thus,  in  four-place  tables  not  beyond  1  to  8686, 
in  five-place  tables  not  beyond  1  to  86859, 
in  six-place  tables  not  beyond  1  to  868599, 

and  so  on. 

It  is  thus  apparent  that  the  division  should  never  be 
carried  in  such  tables  beyond  two  places  ;  generally  not 
beyond  one  place. 

It  is  apparent  from  what  precedes  that  in  a  five-place  table, 
as  soon  as  a  number  rises  above  86859  (the  double  modulus) 
one  in  the  fifth  place  of  the  number  does  not  affect  the 
logarithm  in  the  first  five  figures.  Thus  in  interpolating 
for  numbers  corresponding  to  logarithms  in  the  part  of  the 
table  above  the  double  modulus,  it  is  not  possible  to  get 
the  fifth  figure  within  a  unit  of  accuracy.  In  that  part  of 
the  table  no  interpolation  for  the  fifth  figure  should  be  made. 


28  PLANE   TRIGONOMETRY.  [§  13  b 

So  in  general  for  any  place  table,  we  should  not  interpolate 
for  figures  in  the  place  of  the  table,  when  the  logarithms 
indicate  a  number  beyond  twice  the  modulus. 

EXERCISES. 

1.  Look  up  the  logarithms  of  234  and  235  in  the  four-place  table,  and 
from  them  find  the  logarithms  of  23.48,  2.341,  2.346,  23.49,  234.5,  0.2341. 

2.  Look  up  in  a  five-place  table  the  logarithms  of  4357  and  4358,  and 
from  them  determine  the  logarithms  of  43.572,  4.3578,  432.56,  43579. 

3.  "Write  five  logarithms  at  random,  and  find  in  the  tables  the  num- 
bers corresponding  to  them.  Take  four-place  logarithms  and  five-place 
logarithms. 

4.  Examine  your  table,  and  see  at  what  part  of  the  table  the  tabular 
differences  are  largest. 

5.  Take  any  two  consecutive  numbers  greater  than  twice  the  modulus 
and  having  as  many  figures  as  the  place  of  the  table,  and  examine  as  to 
whether  in  the  table  they  have  the  same  logarithms  to  the  place  of  the 
table. 

6.  The  teacher  may  exercise  the  class  at  will  on  examples  like  Exs. 
1,  2,  3,  so  selecting  them  that  the  first  part  of  the  table,  the  middle  of 
the  table,  and  the  part  of  the  table  above  the  double  modulus  are  all 
considered,  until  the  class  is  thoroughly  familiar  with  how  to  interpolate, 
when  to  interpolate,  how  far  to  interpolate,  and  when  not  to  interpolate. 

§  14.    What  Place  Table  to  Use. 

When  we  have  before  us  a  five-place  table,  we  may  con- 
sider that  it  has  been  made  from  a  six-place  table  by  dropping 
the  sixth  figure,  following  the  usual  rule  as  to  the  final 
figure  as  compared  with  5  (see  §  13,  a,  Note).  In  any 
particular  logarithm  we  do  not  know,  then,  when  we  have 
a  five-place  table,  what  the  sixth  figure  of  the  logarithm  is. 
The  logarithm  may  be  in  error  by  almost  5  in  the  sixth  place, 
in  either  direction.  We  must  thus  allow  in  any  given  loga- 
rithm of  five  places  a  possible  error  of  almost  ±  0.000005. 
In  §  13  we  found 

e  __  log-difference 
x  modulus 


§  14]  DISCUSSION  OF  LOGARITHMS.  29 

If,  then,  there  is  an  error  of  ±0.000005  in  a  logarithm,  the 

corresponding  value  of  -,  or  the  relative  error  in  the  corre- 
sponding number,  is 

±0.000005        ±1 


0.43429  . . .      86859 


,  nearly. 


Thus,  if  we  have  a  single  logarithm  and  we  are  certain  that 
it  is  correct  to  five  places,  the  corresponding  number  can  be 
determined  to  within  one  86859th  part  of  its  value.  This  is 
far  beyond  the  degree  of  accuracy  with  which  measurements 
in  engineering  work  are  usually  carried  out. 

Similarly,  if  a  logarithm  is  known  accurately  to  four  places, 
the  corresponding  number  can  be  determined  to  within  one 
8686th  of  its  true  value.  Hence,  for  a  great  many  engineer- 
ing operations  a  four-place  table  is  sufficient. 

It  must  be  stressed  here  that  we  say,  If  the  logarithm  is 
known  to  be  correct  to  five  figures,  or  four  figures. 

If  the  given  logarithm  has  itself  resulted  from  a  manipula- 
tion of  values  themselves  uncertain  in  the  next  place  beyond 
their  given  place,  the  resulting  logarithm  may  itself  be  uncer- 
tain in  one  or  more  of  its  terminal  places. 

The  effect  of  such  uncertainty  is  considered  in  the  next 
chapter,  so  far  as  it  relates  to  ordinary  trigonometrical  cal- 
culations. 

Later  it  will  appear  that  if  the  sides  of  a  diagram  show 
only  one  significant  figure,  or  only  two,  or  only  three,  or  only 
four,  the  angles  should  not  show  seconds,  and  a  four-place 
table  is  sufficient.  When  the  angles  show  seconds,  a  five- 
place  table  is  called  for.  When  the  angles  show  tenths  of 
seconds,  a  six-place  table  is  called  for ;  hundredths  of  seconds, 
a  seven-place  table. 

An  extension  of  the  table  by  one  place  implies,  as  is  clear 
from  what  precedes,  a  diminution  of  allowable  error  in  read- 
ings and  in  the  results  of  calculation  by  one-tenth,  as  also  in 
the  data. 

A  diagram  with  its  sides  reading  to  one,  two,  or  three 
significant   figures   and   showing   seconds   in   angles  would 


30  PLANE   TRIGONOMETRY.  [§  14 

indicate  that  angles  had  been  measured  more  carefully  than 
lines,  and  would  thus  be  absurd. 

When  we  use  a  table  of  more  places  than  the  significant 
figures  of  the  data  call  for,  we  must  not  take  calculated  re- 
sults to  the  degree  of  accuracy  of  the  table,  but  cut  them 
back  to  the  pattern  of  the  data.  This  will  be  made  clear  in 
the  next  chapter. 

EXERCISES. 

Calculate  to  four  significant  figures  the  value  of  the  following : 


1.  27.34  x  13.56.  6.   2^12. 


2.   2.374  x  0.0732.  7. 


V3lM 
71.36  x  21.27 
3763  x  0.003721' 


3.314.3X0.3164.  8 


V0.003456  x  \/56.73 
4.   27.31  x  273.1  x  0.003416.  9.    (26.31) 2141  x  0.3465. 


316.1 
21.32* 


10.   (2.341) (L361>. 


Calculate  to  five  significant  figures  the  following: 

7.8321  x  0.032564 


11.  73.214  x  3.2154.  14. 


2.3146  x  0.056378 


12.   384.62  x  2.7184.  15.    V7U856  x  V7\3214. 


13    56.732  x  87.563  16       V0.0035678  x  V0.043785 


7-2134  ^0.0073214  x  vU00032156 

If  there  be  more  than  one  calculator,  what  is  the  best  plan  for 
checking  such  calculations  as  the  above  ?  What,  if  there  be  only  one 
calculator  ? 

§  15.    Negative  Mantissa. 

It  occasionally  happens  that  a  calculation  can  be  carried 
out  more  readily  by  having  the  logarithms  that  are  usually 
part  positive  and  part  negative  all  negative,  as,  for  example, 
in  getting  the  value  of 

z  =  (0.00347)00567, 

and  similar  expressions. 


§  15J  DISCUSSION   OF   LOGARITHMS.  31 

The  usual  logarithms  would  be 

log  x  =  0.0567  log  (0.00347) 

=  0.0567x3.54033. 

.-.  log  (log  x)  =  log  (0.0567)  +  log  (3.54033). 

The  last  number  is  part  positive  (mantissa)  and  part  nega- 
tive (characteristic).  Getting  its  logarithm  in  this  form  is 
not  possible.     Instead,  we  use 

log  (log  x)  =  log  (0.0567)  +  log  (2.45967)  (n) 

=  log  (0.0567)  +  log  (2.460)  (n) 

=  2.75358  +  0.39094  (n) 

=  1.14452  (*) 

.-.  logs  =  -0.13948 

or  log  x  =  1.86052 

.\  a;  =  0.7253. 

In  similar  expressions  of  the  form 

y  =  a"\ 

it  is  best  to  calculate  the  value  of  log  (£>c)  =  log  a?,  as  in  the 
preceding  example,  for 

log  y  =  x  log  a, 
and  log  log  y  =  log  x  +  log  log  a. 


EXERCISES. 
Find  the  values  of 

1.  (0.003468)°  o04378  to  four  significant  figures. 

2.  (0.0056143)°  o0036213  to  five  significant  figures. 

3.  (0.3468) (0003214)00003614  to  four  significant  figures. 


4-   (  ^0.0346) ^0.347)  °-361  to  three  significant  figures. 


CHAPTER  II. 

CALCULATION    VICES    AND    DEVICES. 

[Note.  —  When  to  stop  "  figuring  "  is  what  the  student  must  know 
in  calculations.  In  Trigonometry  is  the  first  place  where  the  mature 
student  faces  this  question.  This  chapter  is  intended  as  a  preparation 
for  judicious  calculation.] 

§  16.  When  a  calculation  involving  only  multiplications  and 
divisions  is  to  be  carried  out,  one  may 

(1)  Do  the  work  as  in  arithmetic. 

(2)  Shorten  the  arithmetic  process  by  dropping  the  figures 

which  fall,  in  the  final  result,  beyond  the  place 
(decimal  or  integer)  of  possible  accuracy.  (See 
§  26  and  §  40,  for  the  shortened  process  of  multipli- 
cation and  division.) 

(3)  Use  logarithms. 

(4)  Use  a  sliding-rule  (which  mechanically  adds  and  sub- 

stracts  logarithms  and  indicates  the  corresponding 
number).* 

(5)  Use  a  calculating  machine  designed  for  office  work. 

Which  process  to  use  depends  upon  the  extent  of  work  to  be  done. 
A  single  multiplication,  like  83  times  72,  or  even  like  217.3  times  31.46, 
can  be  carried  out  more  quickly,  perhaps,  directly  than  by  logarithms, 
since,  in  the  latter  case,  one  must  find  the  table,  then  look  up  the 
logarithm  of  each  number,  and  then  look  up  the  corresponding  number. 
However,  if  more  than  three  operations  are  necessary,  and  the  numbers 

*  Every  engineer  who  has  much  calculating  to  do,  will  find  a  pocket 
sliding-rule  of  great  convenience.  The  makers  furnish  an  explanatory 
pamphlet.  A  description  of  the  theory  on  which  they  are  based  will  also 
be  found  in  Raymond's  "Surveying." 

32 


§  18]  CALCULATION  VICES   AND  DEVICES.  33 

are  not  small,  there  is  a  great  saving  of  time  in  using  logarithms. 
Speed,  however,  in  using  logarithms,  as  in  any  other  labor-saving 
device,  depends  on  a  thorough  acquaintance,  from  long  practice,  with 
the  process. 

§  17.     The  numbers  occurring  in  calculations  may  be 

(1)  Known  exactly. 

(2)  Known  only  approximately. 

(3)  Known  exactly,  but  used  only  approximately. 

Since  most  of  the  numbers  used  in  calculations  in  engineering 
work  come  from  measurements  which  are  of  necessity  made  only  ap- 
proximately, or  else  come  from  other  calculations  carried  out  only 
approximately,  we  may  make  the  broad  statement  that  an  engineer  is 
concerned  chiefly,  in  calculations,  with  approximate  numbers.  The  chief 
vice,  then,  of  the  engineering  student  (we  might  also  include  many  engi- 
neers), as  a  calculator,  is  a  more  or  less  total  disregard  of  the  fact  that 
the  numbers  used  are  only  approximate,  with  the  consequent  super- 
abundant care  in  the  extent  of  "  figuring  "  done.  His  vice  is  "  cipher- 
ing "  gone  mad. 

§  18.   Significant  Figures. 

All  figures,  other  than  zero,  are  significant  in  a  number. 
A  zero  is  significant  unless  used  merely  to  locate  the  decimal 
point.  In  207,  2.07,  or  0.207,  the  central  zero  is  significant. 
A  zero  used  merely  to  locate  the  decimal  point  is  not  signifi- 
cant. Thus  in  0.0003,  the  zeros  are  not  significant,  for  an 
error  in  the  next  place  to  the  right  of  3  would  be  the  same 
relative  error  in  0.0003  as  in  tenths'  place  for  3.  Thus  0.0003 
is  not  called  a  number  of  four  significant  figures,  but  of  one 
significant  figure.  A  final  zero  may  be  quite  significant. 
In  such  cases  it  should  always  be  retained.  Thus  if  a  line 
measured  to  the  nearest  hundredth  of  an  inch  is  found  nearer 
29.30  than  to  29.31  or  to  29.29,  the  result  should  be  written 
29.30  and  not  29.3,  for  the  latter  would  mean  that  the  measure- 
ment had  been  made  only  to  the  nearest  tenth.  Frequently, 
however,  ending  zeros  are  not  significant.  For  example, 
if  it  is  said  that  a  certain  building  cost  about  $35,000, 
evidently  the  three  zeros  are  not  significant.    When  the  con- 


34  PLANE   TRIGONOMETRY.  [§18 

text  does  not  indicate  the  degree  of  inaccuracy  of  an  approxi- 
mate number,  there  should  be  some  accompanying  statement 
to  make  the  matter  clear.  Frequently  a  number  is  written 
in  the  form  35.246,  ±  0.012,  with  the  understanding  that  the 
error  may  be  as  great  in  either  direction  as  0.012,  and  that, 
consequently,  the  true  value  lies  somewhere  between  35.258 
and  35.234.  The  same  notation  is  used  in  Least  Squares  in 
an  entirely  different  sense,  namely,  that  it  is  an  even  chance 
that  the  error  in  the  result  is  larger  or  smaller  than  the 
indicated  error.  The  inaccuracy  to  which  a  result  is  liable 
may  also  be  expressed  as  a  per  cent.  Thus  217,  ±  2.17, 
means  the  same  as  217  with  possible  inaccuracy  of  1  fo.  A 
statement  like  31.145693,  ±  0.032,  would  be  absurd,  as  the 
last  three  figures,  693,  would  then  be  meaningless.  This 
should  be  written,  31.146,  ±  0.032.  With  the  exception 
already  mentioned  in  the  case  of  zero,  all  figures  in  a  number 
are  significant,  that  is,  any  change  in  the  figures  would  indi- 
cate a  different  number. 

§  19.   Rejection  Error. 

Approximate  numbers  of  very  frequent  occurrence  are  : 
logarithms ;  square  roots  of  numbers  like  2,  7,  etc. ;  numbers 
like  7r  =  3.14159  .  .  .  and  e  =  2.71  .  .  . ;  numbers  like  those 
treated  later  in  this  book  and  called  sines,  cosines,  etc., 
which,  with  a  few  exceptions,  belong  to  the  large  class  of 
non-terminating,  non-ending  decimals.  Approximate  values 
of  such  numbers  are  obtained  by  stopping  at  any  place, 
increasing  the  figure  in  that  place  by  1  when  it  is  followed 
by  a  5  or  by  a  figure  larger  than  5,  —  otherwise,  leaving  the 
final  figure  unchanged.  Thus,  approximate  values  of  it  are 
3.1,  3.14,  3.142,  3.1416.  Such  approximations  are  always 
nearer  the  true  value  than  5  in  the  next  place  to  the  right. 
Thus  3.14  is  too  small  for  tt,  but  not  by  0.005  ;  3.142  is  too 
large,  but  not  by  0.0005.  The  errors  introduced  into  calcu- 
lations by  thus  cutting  the  number  of  figures  in  a  number 
are  called  rejection  errors.     The  limit  for  the  extreme  value 


§20]  CALCULATION   VICES  AND   DEVICES.  35 

of  the  rejection  error  in  any  approximate  number  entering 
directly  into  a  calculation  is  5  in  the  next  place  to  the  right  of 
the  ending  place.  This  error  in  the  elements  of  a  calculation 
can  effect  very  materially  the  final  result  of  the  calculation. 

§  20.    Effect  of  Errors  in  the  Terms  of  an  Addition-subtraction 

Result. 

If  a  is  liable  to  an  error  xv 

and  b  is  liable  to  an  error  xv 

evidently  a  ±  b  is  liable  to  the  error  ±  (xx  +  #2)>  for  the  error 
in  a  and  b  may  both  chance  to  lie  the  same  way. 

Similarly,  for  any  number  of  such  operations,  the  largest 
possible  error  is  ±  (xx  -f-  x2  +  xz  +  x4  -f-  •  •  •  +  #„)• 

It  is  extremely  unlikely,  of  course,  that  in  a  very  large 
number  of  such  terms,  taken  at  random,  where  the  error 
in  each  term  is  as  likely  to  lie  one»way  as  the  other,  the 
errors  in  all  the  terms  will  lie  the  same  way.  Experience 
has  shown  that  in  such  cases  there  is  a  balancing  of  errors. 
When,  however,  the  number  of  terms  is  small,  it  is  not  at 
all  uncommon  to  find  the  errors  all  running  one  way.  The 
student  can  give  himself  some  acquaintance  with  this  matter 
by  selecting  logarithms,  in  groups  of  two,  of  three,  of  four  — 
as  will  be  the  case  in  most  of  the  calculations  in  trigonom- 
etry —  at  random,  from  a  four-place  table,  and  afterwards 
looking  up  the  fifth  figure  in  a  five-place  table,  or  higher  place 
table.  The  errors  will  be  far  from  balancing.  It  is  quite 
a  common  error  to  apply  the  deductions  of  the  Method  of 
Least  Squares,  based  as  they  are  on  "the  long  run,"  to  a 
very  short  run. 

A  not  uncommon  disregard  of  the  preceding  deductions  is 
shown  in  adding  (subtracting)  approximate  numbers. 

For  example,  in  the  addition, 

27.31 
346.2159 
373.5259' 


36  PLANE  TRIGONOMETRY.  [§20 

if  these  numbers  are  liable  to  rejection  error,  the  final  result 
is  quite  misleading,  for  it  makes  it  appear  that  the  sum  is 
liable  only  to  the  error  0.00005,  whereas  it  is  0.00505.  In 
such  cases  the  number  with  many  decimal  places  should  be 
cut  back  to  the  pattern  of  the  other. 

Thus,  27.31 
346.22 
373.53,  where  the  final  3  may  be  "off"  by  1. 

In  additions  and  subtractions  of  approximate  numbers,  subject 
to  rejection  error,  let  each  term  show  the  same  number  of  decimal 
places. 

§  21.    Applications  of  the  Suggestions  of  §  20. 

If  the  number  of  terms  of  an  addition-subtraction  expres- 
sion is  n,  and  each  term  is  liable  to  rejection  error  beyond 
the  rth  decimal  place,  the  extreme  limit  for  the  accumulation- 
effect  of  these  rejection  errors  in  the  algebraic  sum,  is  5  n  in 
the  (r  +  l)th  place.  Suppose  20  terms  lead  to  the  result, 
1000,  the  smallest  sequence  of  four  significant  figures,  with 
20  times  5  in  tenths'  place  as  extreme  accumulation  rejection 
error,  or  10.  This  is  1 JG  of  the  sum.  If  the  result  had  been 
9999,  the  largest  four-figured  number,  the  accumulation 
rejection  error  would  still  be  10,  but  as  a  per  cent,  only  a 
little  more  than  one-tenth  of  1  fo.  Similarly,  for  10000  and 
99999,  the  smallest  and  largest  five-figured  numbers,  the  per 
cents  are  ^  of  1  fo  and  jfa  of  1  fo. 

Since  per  cents  are  independent  of  the  position  of  the 
decimal  point,  the  following  computation  rule  is  sometimes 
advised : 

If  an  addition-subtraction  expression  is  desired  within  the 
range  of  -^  of  1%  to  1%  possible  inaccuracy,  let  the  final 
result  show  four  significant  figures.  If  the  permissible  error 
is  to  lie  between  -^  of  1%  and  y^  of  1%,  let  the  final  result 
show  five  significant  figures. 

This  rule,  deduced  as  just  shown,  takes  the  worst  cases 
and  the  best  cases,  presuming  in  each  case  that  the  effect  of 


§21]  CALCULATION  VICES  AND  DEVICES.  37 

rejection  error  lies  in  each  term  all  the  time  the  same  way, 
and  with  the  rather  unusual  number  of  20  terms.  The 
"  factor  of  safety  "  is  thus  so  great  that  in  the  case  of  2,  3, 
4,  5,  6  terms,  which  are  far  more  common  than  numbers  of 
terms  very  much  larger,  it  is  very  much  better  to  disregard 
the  rule  and  take  the  number  of  figures  to  suit  the  individual 
example,  as  illustrated  in  the  following  examples. 

EXERCISES. 

1.  Find  to  within  0.4  of  1  %  the  value  of 

55.6312  +  71.8371  -  4.32713. 

By  inspection,  the  result  is  about  120,  of  which  0.4  of  1  %  is  about  0.5- 
Since  there  are  only  three  terms,  the  maximum  accumulation  rejec- 
tion error  is  0.15,  if  hundredths  are  dropped.  This  is  within  the  limit 
required.     Therefore  as  follows  : 

55.6  +  71.8  -  4.3  =  123.1,  ±  0.15. 

2.  Find  to  within  0.6  %  the  value  of 

47.3489  +  174.32825  -  5.62147. 

By  inspection,  the  result  is  about  200,  of  which  0.6  %  is  1.2.  Thus, 
hundredths  may  be  dropped,  and  we  have 

43.3  +  174.3  -  5.6  =  216.0,  ±  0.15. 

3.  Make  up  and  solve  some  examples  like  the  preceding,  some  with 
three  terms,  some  with  four,  some  with  five,  six,  seven  terms. 

4.  Add  273.1415  and  564.1.     To  what  error  is  the  sum  liable  ? 

5.  Add  32.14156,  32.718,  32.6.     To  what  error  is  the  sum  liable  ? 

6.  Find  the  value  of  314.2156783  +  315.61  -  2.3124  -  87.4. 

7.  Find  the  sum  of  36.732,  ±  0.021,  and  71.2468,  ±  0.0004. 

The  uncertainty  in  the  first  number  being  the  greater,  that  decides 
the  number  of  decimal  places  to  retain  in  the  second.  Siuce  0.0004 
added  to  the  final  8  of  the  second  number  makes  the  6  uncertain  bjy  1, 

we  may  use 

36.732  ±  0.021 
71.246  +  0.001 


107.978  ±  0.022 
the  negative  limit  being  not  quite  this. 
8.   Add  3.607(±  0.022)  and  5.84(±  0.28). 


38  PLANE   TRIGONOMETRY.  [§22 

9.   Reduce  to  a  single  term 

7.0845(±  0.0012)+  7.364(±  0.001)-  3.2748(±  0.012). 
10.   Find  the  value  of 

53.3456(±  0.05%)+  167.3578(±  0.01%)-  6.2315(±  0.03%). 

§  22.    Application  to  Logarithmic  Work. 

The  most  important  practical  application  of  the  results 
reached  in  the  preceding  sections  concerning  the  effect  of 
accumulation  rejection  error  is  in  the  use  of  logarithms. 
The  logarithm  of  a  product  is  the  sum  of  the  logarithms  of 
its  factors,  etc. 

Take  the  case  of  two  logarithms  added  (subtracted). 
Consider  two  four-place  logarithms  as  a  special  case.  The 
result  may  possibly  be  uncertain,  in  the  extreme  case,  by 
almost  1  in  the  fourth  place.  If  ten  such  logarithms  are  added 
(subtracted),  the  result,  in  the  extreme  case,  may  be  uncer- 
tain by  5  in  the  fourth  place,  so  that  the  resulting  logarithm 
has  only  the  accuracy  of  a  logarithm  taken  from  a  three-place 
table.  These  simple  facts  are  constantly  disregarded  in  the 
use  of  four-place  tables,  as,  for  example,  when  one  interpo- 
lates to  get  a  fifth  figure  in  a  number  corresponding  to  a  log- 
arithm obtained  by  additions  (subtractions)  of  logarithms. 

It  has  been  pointed  out  in  §  12,  page  24,  that  if  two 
logarithms  differ  by  the  small  quantity  e,  the  ratio  of  the 
difference  of  the  two  corresponding  numbers  to  the  smaller 
number    (the   "relative   error"   in    the    two    numbers)   is 

T~, —  or  „  ,     * In  constructing  a  four-place  table, 

modulus       0.4342  ...  5  F 

0.00014  counts  as  1  in  the  fourth  place,  while  0.00015  counts 
as  2,  as  would  also  0.00024.  Thus  the  "  relative  error"  corre- 
sponding to  the  accumulation  rejection  error  1  in  the  fourth 
place,  in  the  case  of  an  addition  (subtraction)  of  two  four- 

,        ,  t     0.00014  ,12  3 

p  ace  logs  may  be  or  about  —  or  _  or  _, 

etc. 

Thus  if  two  four-place  logs  are  added,  giving  a  log  cor- 
responding to  a  number  lying  between  3100  and  6200  (the 


§23]  CALCULATION   VICES  AND   DEVICES.  39 

decimal  point  falling  where  it  may),  the  rejection  error  of 
1  in  the  fourth  place  will  leave  the  corresponding  number 
in  doubt  as  to  three  consecutive  numbers  (and  of  course  all 
intermediate  numbers).  If  the  corresponding  number  lies 
between  6200  and  9300,  it  will  be  in  doubt  among  five  con- 
secutive numbers  and  the  intermediates.  It  will  be  in  doubt 
among  seven  consecutive  numbers  and  their  intermediates 
when  it  lies  above  9300. 

From  this  one  will  readily  draw  the  conclusion  that  in 
using  a  four-place  table  one  ought  not,  as  a  rule,  to  interpo- 
late for  a  fifth  figure,  and  one  sees  that  the  fourth  figure  is 
in  doubt  as  soon  as  the  number  corresponding  to  a  given  log 
which  has  arisen  by  adding  two  logs,  or  subtracting  them, 
rises  above  3100.  This  can  be  observed  in  a  four-place 
table. 

EXERCISES. 

1.  Two  logs  added  (subtracted)  give  2.6042 ;  what  is  the  correspond- 
ing number? 

Ans.     401.9  or  402.0  or  402.1  or  any  intermediate  number. 

2.  Two  logs  added  (subtracted)  give  2.8482 ;  what  is  the  correspond- 
ing number  ? 

Ans.    704.8  or  704.9  or  705.0  or  705.1  or  705.2  or  any  number  inter- 
mediate to  these. 

3.  Two  logs  added  (subtracted)  give  3.9845 ;  what  is  the  correspond- 
ing number? 

Ans.    9647;  9648;  9649;  9650;  9651;   9652;  9653;  or  any  number 
intermediate  to  these. 


§  23.    Conclusion  as  to  What  Place  Table  to  Use. 

From  the  preceding  section  it  will  be  clear  that  when 
the  number  of  additions  (subtractions)  of  logs  is  few,  and 
where,  consequently,  a  balancing  of  errors  is  not  to  be 
counted  on,  one  ought  to  use  a  table  giving  at  least  one 
figure  beyond  the  number  of  figures  desired  in  the  calculated 
results.     In  20  additions  with  a  seven-place  table,  it  may 


40  PLANE   TRIGONOMETRY.  [§24 

happen  that  the  fifth  figure  is  in  error  by  1.  If  the  charac- 
teristic calls  for  more  than  four  figures  in  using  a  four-place 
table,  the  figures  following  the  fourth  should  be  filled  with 
non-significant  zeros.  Such  a  characteristic  indicates  that  a 
four-place  table  ought  not  to  be  used. 

§  24.    Application  to  Some  of  the  Problems  of  Trigonometry. 

1.  It  will  be  found  later  in  solving  right-angled  triangles 
that  two  logs  are  added  (subtracted).  The  effect  is,  there- 
fore, as  pointed  out  in  §  22. 

2.  In  solving  oblique-angled  triangles,  except  when  three 
sides  are  given,  it  will  be  found  later  that  three  logs  are 
united.  In  using  a  four-place  table,  the  fourth  place 
becomes  doubtful  by  1.5.  This  counts  as  2  in  the  table 
construction.     So  does  2.4. 

"  Relative  error  "  =  — ■ = = = = 

0.4342  .  .  .     1770     3500     5300     7000 

,  as  approximate  results. 


8900 

Consequently,  in  such  work,  if  the  resulting  logarithm 
corresponds  to  a  number  (irrespective  of  decimal  point) 
which  lies 

between  1770  and  3500,  there  is  doubt  among  three  consecutives, 
between  3500  and  5300,  there  is  doubt  among  five  consecutives, 
between  5300  and  7000,  there  is  doubt  among  seven  consecutives, 
between  7000  and  8900,  there  is  doubt  among  nine    consecutives, 

and  all  intermediate  numbers. 

EXERCISES. 

If  three  logs  added  (subtracted)  give  the  following  logs,  state  the 
corresponding  numbers : 

1.   2.2989.         2.   1.5955.         3.   2.7404.         4.   1.8633.         5.   3.9294. 

3.  In  solving  a  triangle  when  the  three  sides  are  given,  it 
will  be  found  later  that  four  logs  are  added  (subtracted), 
and  this  result  is  divided  by  2  on  account  of  a  square  root. 
The  resulting  doubt  in  the  fourth  place  is  therefore  1,  and 
this  case  is  the  same  as  for  solving  right-angled  triangles. 


§25c]  CALCULATION  VICES  AND  DEVICES.  41 

§  25.   Errors  in  Products  of  Two  Factors. 

(a)  If  a  is  liable  to  an  error  ±  xv  and  b  is  liable  to  an 
error  ±  rr2,  to  find  the  error  to  which  the  product  ab  is  liable. 

Let  av  bv  be  the  true  values  of  a,  b, 

a1  =  a  ±  £j, 

b1  =  b  ±  xv 
axbx  =  ab  ±  ax2  ±  bxx  +  xxx2. 
If  xv  x2  are  small,  xxx2  may  be  neglected. 
,\  error  in  ab  is  ±  (ax2  -f  taj), 
or,  a  times  b's  error  plus  b  times  d's  error. 

(If)  If  a  is  liable  to  an  error  of  p1  per  cent,  and  b  is  liable 
to  an  error  of  p2  per  cent,  to  what  per  cent  error  is  the 
product  ab  liable  ? 


Here, 


.   _  aPi 
1      100' 


and  **  =  m' 

ax2+bx1  =  j^(p1+p2). 

.\  per  cent  error  in  the  product  ab  is  the  sum  of  the  per 
cents  for  the  factors. 

This  fact  is  frequently  stated  in  a  manner  quite  misleading, 
namely,  "A  product  cannot  be  more  accurate  than  its  least 
accurate  factor."  Since  when  the  errors  are  unknown  the 
worst  case  must  be  allowed  for,  it  is  not  how  small  the  error 
may  be,  but  how  large,  that  is  the  important  matter.  If  a  is 
liable  to  2<fo  error  and  b  to  3jfc  error,  ab  is  liable  to  5jfc  error. 
It  is  not  sufficient  to  say  that  ab  cannot  be  more  accurate 
than  b. 

(<?)  If  xv  x2  are  rejection  errors,  to  find  the  accumulation 
rejection  error  in  ab,  let 

a  = ,  and  o  = , 

(10)*  (10)* 


42  PLANE  TRIGONOMETRY.  [§25c 

which  signifies  that  A,  B  are  the  numbers  a,  b,  when  the 
decimal  point  is  disregarded,  and  dv  d2  are  the  numbers  of 
decimal  places  in  a,  b  respectively. 

xx  is,  at  most,  5  in  the  (c?x  +  l)th  decimal  place, 
that  is,     xx  is,  at  most,  J  in  the  c^th  decimal  place, 
and  x2  is,  at  most,  J  in  the  c?2th  decimal  place. 

2    (10)*+* 

.-.  to  get  the  extreme  accumulation  rejection  error  in  the 
product  ab,  add  a  and  b  without  regard  to  decimal  point,  and 
then  point  off  a  number  of  decimal  places  equal  to  the  sum  of  the 
decimal  places  in  a  and  b  ;  then  divide  by  2. 


"Relative  error"  in  ab 


•   1/1  ,   V\ 

1S2U  +  5/ 


EXERCISE. 

What  is  the  extreme  accumulation  rejection  error  in  3.142  times 
36.52? 

Solution. 

3142 

3652 

2).06794 

.03397 

Consequently  the  hundredths'  place  in  the  product  is  uncertain  by 
more  than  3. 

When  such  a  product  is  formed  by  the  usual  arithmetic 
process,  a  large  part  of  the  "  ciphering"  is  a  waste  of  time, 
since  figures  beyond  the  place  of  doubt  are  written  down. 
In  forming  a  product  like  that  just  given,  the  numbers  being 
subject  to  rejection  error,  writing  figures  in  places  to  the 
right  of  hundredths  is  an  absurdity.  The  actual  multiplica- 
tion, when  logs  are  not  used,  should  be  carried  out  by  the 
method  of  §  26.  When  logs  are  used,  the  result  should  be 
given  only  as  far  as  to  the  doubtful  place. 


§26]  CALCULATION   VICES  AND   DEVICES.  43 

§  26.    The  Shortened  Process  of  Multiplication. 

First,  determine  the  uncertain  place  in  the  product,  as 
above.  Set  the  numbers  so  that  decimal  points  fall  one 
under  the  other.  In  the  product,  set  down  no  figures  to  the 
right  of  the  doubtful  place.  Perform  no  multiplications  two 
or  more  places  to  the  right  of  the  doubtful  place.  Do  this 
only  one  place  to  the  right  of  the  doubtful  place,  but  set  down 
no  results  ;  use  such  products  only  to  "  carry  "  to  the  doubtful 
place.  From  a  product  like  27  "  carry  "  3  (an  extra  1,  because 
7  is  more  than  5) ;  the  same  for  25 ;  for  24,  carry  2,  and  so  on. 
Carry  out  the  multiplication  from  left  to  right  on  the  multi- 
plier, instead  of  from  right  to  left,  as  in  arithmetic.  Start 
the  multiplication  with  that  figure  of  the  multiplicand  which 
gives  a  figure  one  place  to  the  right  of  the  doubtful  place. 
As  one  moves  to  the  right  one  figure  each  time  on  the  multi- 
plier, start  the  multiplication  one  figure  to  the  left  on  the 
multiplicand  with  each  such  move  along  the  multiplier. 

ILLUSTRATIVE  EXAMPLES. 

Find  3.142  x  3G.53.     As  already  shown,   hundredths  are  doubtful 

by  0.03. 

„      .,  .    ,  .       ,  By  shortened  process 

By  old-fashioned  process.  (hundredths  doubtful). 

36.53  36.53 

3.142  3.142 

7306  109.59 

14612  3.65 

3653  1.46 

10959  7 

114.77726  114.77  ±0.03 

Other  examples  by  the  shortened  process : 


its  doubtful. 

Tenths  doubtful. 

Hundredths  doubtful. 

43.29 

27.314 

96.789 

3.8 

71.65 

21.579 

130. 

1912.0 

1935.78 

34. 

27.3 

96.79 

164. 

±2 

16.4 

48.39 

1.4 

6.77 

1957.1  ±0.2 

.86 

2088.59  ±0.06 


44  PLANE  TRIGONOMETRY.  [§26 

In  the  first  example,  since  hundredths  are  doubtful,  the 
multiplication  is  started  by  using  3  of  the  multiplier  with 
the  3  on  the  right  of  the  multiplicand.  In  taking  next  the 
1  of  the  multiplier,  the  multiplication  is  started  on  the  mul- 
tiplicand with  the  5,  —  motion  to  the  right  on  the  multiplier 
with  motion  to  the  left  on  the  multiplicand.  There  is  in 
this  line  no  carrying  from  the  1  times  3,  since  this  is  less 
than  5.  In  the  next  line,  where  4  of  the  multiplier  is  used, 
the  multiplication  is  started  with  the  6  of  the  multiplicand, 
there  being  a  "  carrying  "  of  2  from  the  4  times  5,  one  place 
to  the  right  of  the  doubtful  place.  In  the  next  line,  where 
the  2  of  the  multiplier  is  used,  the  start  is  on  the  left-hand 
3  of  the  multiplicand,  with  1  carried  from  the  2  times  6, 
this  being  one  place  to  the  right  of  the  doubtful  place. 

In  the  second  example,  where  units  are  doubtful,  the  start 
is  made  by  3  times  the  unit  3  of  the  multiplicand,  with  1 
carried  from  3  times  the  2  to  the  left  of  this  figure. 

Where  to  start  the  multiplication  on  the  multiplicand  is 
easily  seen.  Consider  the  third  example,  where  tenths  are 
doubtful ;  the  initial  7  of  the  multiplier  is  70.  If  70  is  mul- 
tiplied by  the  4  on  the  right  of  the  multiplicand,  this  is  in 
fact  multiplying  by  0.004,  with  the  result  0.28,  and  we 
should  thus  have  the  figure  8  in  hundredths'  place,  whereas 
tenths'  place  is  the  doubtful  place.  Consequently,  the 
multiplication  is  started  by  taking  7  times  the  1  of  the 
multiplicand,  carrying  3  from  the  7  times  the  4,  which  is 
one  place  to  the  right  of  the  doubtful  place.  A  careful 
study  of  the  examples  worked  out  will  make  the  process 
clear. 

The  place  in  the  multiplicand  at  which  to  make  the  initial 
start  is  given  by  the  following  rule : 

Calling  units'  place  the  zeroth  place ;  tenths'  place,  place 
1 ;  hundredths'  place,  place  2,  and  so  on ;  tens'  place  as  place 
—  1 ;  hundreds'  place  as  place  —  2,  and  so  on ;  the  starting 
place  in  the  multiplicand  from  which  to  set  down  figures  is 
given  by  subtracting  from  the  doubtful  place  of  the  product 


§27]  CALCULATION   VICES  AND   DEVICES.  45 

the  place  of  the  extreme  left-hand  figure  of  the  multiplier. 
Carry  from  one  place  to  the  right  of  this. 

Thus,  in  the  first  example,  the  doubtful  place  of  the  prod- 
uct is  2,  and  the  place  of  the  extreme  left-hand  figure  of  the 
multiplier  is  zero.  The  start  is  on  the  place  2  of  the  multi- 
plicand. In  the  last  example,  the  doubtful  place  of  the 
product  is  2,  and  the  place  of  the  left-hand  figure  of  the  mul- 
tiplier is  —  1.     The  start  is  on  place  3  of  the  multiplicand. 

EXERCISES. 

Determine  the  extreme  accumulation  rejection  error  in  the  following 
products.  Find  the  first  place  which  is  uncertain.  Name  the  extent  of 
uncertainty  in  that  place.  Carry  out  the  multiplications  by  the  shortened 
process. 

1.  6.043  times  6.043.  5.  23.57  times  61.23. 

2.  12.65  times  111.7.  6.  0.65  times  0.32. 

3.  78.21  times  1450.  7.  78.21  times  1.45. 

4.  23.57  times  612.3. 

Carry  out  some  of  the  preceding  multiplications  by  the  ordinary  arith- 
metic process,  and  see  if  there  is  any  saving  of  labor.  Carry  out  the  same 
multiplications  by  logs,  and  compare  results.  Make  sure  that  a  log  table 
which  will  give  correctly  the  number  of  figures  in  the  final  products  is 
used,  and  that  the  products  are  not  taken  beyond  the  doubtful  place. 
A  great  deal  can  be  learned  about  a  log  table  from  these  exercises  by 
making  the  comparisons  here  recommended. 

§  27.   To  find  the  Accumulation  Rejection  Error  in  the 
Product  ab  as  a  Per  Cent  Error. 

AB 


(10)«t+d« 
1      A  +  B 


Since  the  product  is 

and  the  error  is 

2     (10)*+*' 

the  per  cent  error  is  [  —  -f  —  ). 

^  2   \A     BJ 

.'.  the  per  cent  accumulation  rejection  error  is  one-half  of  one 
hundred  times  the  sum  of  the  reciprocals  of  a  and  b  when  taken 
without  the  decimal  point. 


46  PLANE   TRIGONOMETRY.  [§27 

EXERCISES. 

1.  To  compare  the  rejection  error  in 

1000  x  1000  and  10.00  x  1.000. 
The  per  cent  error  in  each  product  is 

^(tAv  +  hW).  or  A  of  1%. 

2.  To  compare  the  rejection  error  in 

9999  x  9999  and  99.99  x  9.999. 
The  per  cent  rejection  error  is  in  each  case 

!25M_+   M  or_I_oflo/o. 

2    V9999      9999/         99.99 
that  is,  a  little  more  than  T£7  of  1  %. 

3.  A  comparison  of  Exs.  1  and  2  will  show  that  the  accumulation  re- 
jection error  in  a  single  product  of  two  factors  of  four  significant  figures 
may  vary  from  ^  of  1  %  to  T^  of  1  %  of  the  product. 

4.  Deduce  in  the  same  way  the  corresponding  relations  for  the  product 
of  two  factors  of  five  significant  figures ;  six  significant  figures ;  seven 
significant  figures. 

Summary. 

(1)  For  three  significant  figures,  <     1     >    ^    of  1%. 

(2)  For  four  significant  figures,    <    T^    >  -^o   °f  !%• 

(3)  For  five  significant  figures,     <  T^¥  >  j-^  of  1%. 

(4)  For  six  significant  figures,       <  j-^q  >  -jowo"  of  1%. 

etc.,  etc. 

§  28.    Approximate  Values  for  Per  Cent  Accumulation 
Rejection  Error  in  a  Product  of  Two  Factors. 

Let  the  100  in  (~7  +  ~#)  cance^  tne  two  terminal  fig- 
ures of  A,  B.     Whence  the  rule  : 

(1)  The  per  cent  rejection  error  in  the  product  ab  is  given 
approximately  by  dropping  the  decimal  points  and  the  two 
terminal  figures  of  A,  B,  observing  the  usual  rule  with  ref- 
erence to  the  next  figure  being  5,  more  or  less;  invert;  add ; 
take  half 


§29]  CALCULATION   VICES   AND  DEVICES.  47 

An  approximation  somewhat  rougher,  but  in  many  cases 
close  enough,  is  given  by  the  following : 

(2)  Drop  the  decimal  points  and  two  terminal  figures; 
change  each  remaining  figure,  except  the  initial  figure,  to  zero  ; 
retain  the  initial  figure,  observing  the  usual  rule  with  reference 
to  the  next  figure  being  5,  or  more  or  less  than  5  ;  invert ;  add  ; 
take  half. 

EXERCISES. 

1.  A  square  has  its  side  measured  to  the  nearest  hundredth  inch,  giv- 
ing 25.32.  What  is  the  maximum  per  cent  accumulation  rejection  error 
in  its  area?  Ans.   About  ^z  of  1  %. 

2.  A  rectangle  has  its  sides  measured,  giving  24.32  and  378.2.  What 
is  the  maximum  per  cent  accumulation  rejection  error  in  the  area  ? 

Ans.   About  &  of  1%. 

3.  Apply  the  approximation  rules  to  find  the  maximum  per  cent  re- 
jection error  in  each  product  of  §  26. 


§  29.   Relation  of  the  Two  Factors  of  a  Given  Product  when  the 
Effect  of  Accumulation  Rejection  Error  is  Least. 

„  .  .    100/1   ,   1\         100  (A  +  B\ 

Per  cent  error  is  -y.^  +  jji  or  -g^-^g-J  ■ 

Since  AB  is  fixed  by  hypothesis,  the  per  cent  error  is  least 
when  A  +  B  is  least. 

Let  A=VAB  +  X,  (1) 

B  =  VAB  -  Y.  (2) 

Then  A+B=2VAB  +  X-Y.  (3) 

Also  ( VA  —  VB)2  is  positive,  when  A  is  not  B. 

^  +  jg_2VZ5>0,  or  A  +  B>2VAB. 

.'.  Y  cannot  be  more  than  X  in  (3).  .*.  the  least  value 
which  A  +  B  can  have  in  (3)  is  2  VAB,  which  occurs  when 
A  =  B. 


48  PLANE   TRIGONOMETRY.  [§30 

EXERCISES. 

1.  Compare  the  effect  of  accumulation  rejection  error  in  the  area  of  a 
square  whose  side  is  12.24,  and  the  rectangle  of  about  the  same  area,  its 
sides  being  3.06  and  48.96. 

2.  Compare  the  accumulation  rejection  error  in  (31. 62)2,  and 
9999  x  0.1000,  the  products  being  each  about  999.9000. 

§  30.  To  find  the  Per  Cent  Error  in  the  Product  of  Three  or 
More  Factors,  Each  Factor  being  Liable  to  a  Given  Per  Cent 
Error. 

If  a  is  liable  to  px  per  cent  error,  and  b  to  p2  per  cent 
error,  it  has  been  shown  that  ab  is  liable  to  the  error 
(P1+P2)  Per  cent.  Evidently,  then,  (a6)c  is  liable  to  the 
per  cent  error  CFi+A  +  J's)'  c  being  liable  to  the  error  pz 
per  cent. 

So,  in  general,  the  per  cent  error  to  which  a  product  of 
any  number  of  factors  is  liable,  is  the  sum  of  the  per  cents 
of  the  factors. 

It  is  assumed  here,  of  course,  that  the  absolute  error  in 
each  factor  is  small,  for  only  in  this  case  is  the  error  in  ab 
a  times  5's  error  plus  b  times  a's  error.  In  the  above  deduc- 
tions all  errors  arising  from  product  of  errors  have  been 
neglected. 

The  tendency  of  multiplication  of  factors,  each  subject  to 
error,  is  to  increase  the  maximum  error  to  which  the  product 
may  be  liable.  t 

This  is  a  consequence  of  the  preceding  section,  it  being 
assumed  all  the  while  that  the  worst  possible  case  may  occur 
by  the  errors  chancing  to  fall  all  the  same  way.  When  the 
number  of  factors  is  comparatively  few,  so  that  a  balancing 
of  errors  is  not  to  be  relied  on,  the  tendency  of  multiplication 
is  to  subject  the  result  to  an  increasing  possible  error. 

From  this  it  appears  that  the  measured  quantities  which 
form  the  basis  of  a  calculation  should  be  measured  with  a 
greater  degree  of  accuracy  than  that  expected  in  the  calculated 
results  based  on  them. 

All  measurements  are  subject  to  rejection  error. 


§32]  CALCULATION   VICES  AND   DEVICES.  49 

Later  on,  therefore,  when  the  measured  sides  of  a  diagram 
are  given  to  one  significant  figure,  it  will  be  absurd  to  let 
the  calculated  sides  show  two  or  more  significant  figures. 

The  calculated  parts  of  a  diagram  should  never  show  more 
significant  figures  than  the  measured  parts. 

It  is  hopeless  to  give  calculated  results  seven-place  accuracy 
when  they  are  based  on  field  measurements  with  three-place 
accuracy. 

§  31.     To  find  Numerical  Expressions  for  the  Per  Cent  Accumu- 
lation Rejection  Errors  in  Products  of  Several  Factors. 

For  two  factors  it  is  -rr-f-j  +  -=)« 

Similarly,  for  three  factors  -77- (-7 +  -5 +  7^) 

Similarly,  for  four  factors  — —  (  —  +  —  +  -x,+  —  V 
J  2  \A     B      CD) 

where  A,  B,   (7,  D  are  a,  5,  <?,  d  with  the  decimal  points 
dropped.      (See  §  27). 

Approximate  results  can  be  obtained  by  an  extension  of 
the  two  approximation  rules  given  in  §  28. 

§  32.  On  the  Effect  of  the  Number  of  Significant  Figures  in  the 
Factors  of  a  Product  on  the  Accumulation  Rejection  Error  of 
the  Product. 

From  the  approximation  rules  of  the  preceding  section, 
it  is  apparent  that  if  four  significant  figures  in  each  factor 
give  a  certain  per  cent  rejection  error  in  the  product,  five 
significant  figures  in  each  factor  will,  approximately,  cut 
this  error  to  -fa  of  its  value,  and  three  significant  figures  will 
multiply  it  by  10,  approximately,  and  so  on. 

From  this  it  is  easy  to  determine,  in  any  special  case,  how 
many  significant  figures  to  take  in  each  factor  in  order  to 
secure  a  result  true  to  within  a  specified  per  cent  of  error. 


50  PLANE   TRIGONOMETRY.  [§32 

ILLUSTRATIVE   EXAMPLES. 

1.  Given  tt  =  3.14159,  r  =  8.43276,  h  =  9.76438,  find  the  volume  of  the 
corresponding  cylinder  from  the  formula  it  •  r  •  r  •  h  •,  to  within  1  %. 

Taking  only  integers,  the  per  cent  rejection  will  be  approximately 
H~  ft  +  i  +  I  +  tV)>  or  about  *#*.  The  use  of  any  additional  significant 
figure  practically  divides  this  result  by  10. 

2.  Show  that  if  the  volume  of  a  cylinder  is  gotten  from  tt  =  3.142, 
r  =  6.043,  h  =  12.65,  the  accumulation  rejection  error  is  less  than  -^  of  1%. 

3.  Use  the  shortened  process  of  multiplication  to  find  the  volumes  of 
the  cylinders  in  Exs.  1  and  2,  to  within  1%  and  to  within  ^  0f  1% 
respectively. 

Solution  of  Example  2  to  within  1%. 

Hundredths  doubtful.  Tenths  doubtful.  Doubtful  by  less  than  12. 

6.04  36.5  12.7 

6.04  3.14  115. 

36.48  109.5  1270. 

3.7  127. 

1.4  64. 

114.6  1451. 

The  student  will  find  it  valuable  exercise  to  calculate  the  error  ax2  +  bx1 
for  each  product  as  formed,  and  notice  its  growth. 

4.  What  would  be  the  maximum  effect  of  accumulation  rejection 
error  in  Exs.  1  and  2,  if  we  take 

in  Ex.  1,  tt  =  3.1,  r  =  8.4,  h  =  9.8 ; 

in  Ex.  2,  ir  =  3.1,  r  =  6.0,  A  =  13? 

5.  The  area  of  a  circle  («•  •  r  •  r)  is  obtained  from  ir  =  3.142,  r  =  6.043 ; 
in  what  place  is  the  area  uncertain  from  accumulation  rejection  error  ? 

The  easiest  way  to  answer  a  question  like  this  is  to  determine  the 
maximum  per  cent  rejection  error,  and  from  this  get  the  doubtful  place 
in  the  area  from  a  rough  estimate  of  the  area. 

6.  Determine  the  area  of  the  circle  in  Ex.  5,  using  four  significant 
figures  and  the  shortened  process  of  multiplication,  dropping  in  each 
product,  as  in  the  solution  of  Ex.  2  above,  the  doubtful  place  as  new 
products  are  formed. 

To  within  what  per  cent  will  ir  —  3.1,  r  =  6.0  give  the  area  of  the 
circle  ?     What  if  we  take  tt  =  3  and  r  =  6  ? 

7.  If  in  the  cylinder  of  Ex.  2,  r  =  6.0428,  ±  i  of  1  %  and  h  =  12,653, 
±  jJj  of  1  %  and  7r  =  3.1416,  to  what  per  cent  error  is  the  volume  liable, 
neglecting  the  error  in  tt?  Am.    About  T7^  of  1  %. 

How  many  significant  figures  should  be  taken  in  each  factor  of 
iT'T'T-h  to  get  within  this  error?  Ans.     Four. 


§33]  CALCULATION  VICES  AND  DEVICES.  51 

8.  Consider  the  volume  of  the  cone  r  =  6.18,  h  =  127,  ir  =  3.14,  in  the 
same  manner  as  the  preceding  examples. 

§  33.    On  the  Reasonableness  of  Retaining  the  Same  Number  of 
Significant  Figures  in  the  Factors  of  a  Product. 

Two  measurements  showing  the  same  number  of  significant 
figures  indicate  about  the  same  degree  of  relative  accuracy  of 
measurement.  Thus,  in  36.27  and  4134,  the  former  means 
something  between  36.265  and  36.275,  and  the  latter  some- 
thing between  4133.5  and  4134.5,  so  that  the  largest  relative 
error  in  the  first  is  y^s?  an(l  ^n  the  second  -g-gV'S"'  wlncn  are 
reasonably  near  together.  The  relative  rejection  error  in 
the  first  is  ^g§ y7  and  in  the  second  ^ i|^-q,  or  y^V?  and  -g^o* 

The  widest  possible  discrepancy  occurs,  of  course,  between 
the  largest  and  smallest  sequence  of  the  given  number  of 
places,  as,  for  example,  in  1000  and  9999,  for  four  figures. 

Thus,  unless  some  unusual  reason  can  be  given  why  one 
factor  of  a  product  should  be  measured  with  far  greater 
accuracy  than  others,  all  factors  should  show  the  same  num- 
ber of  significant  figures,  as  a  rule. 

All  the  measured  parts  of  a  diagram  should  thus  show  the 
same  number  of  significant  figures,  and  the  calculated  parts 
should  not  show  more  significant  figures  than  the  measured 
parts.  They  may  show,  generally,  the  same  number  of  sig- 
nificant  figures. 

EXERCISES. 

1.  Find  21746893  x  1.53,  these  representing  measurements. 

Since  the  rejection  error  in  1.53  is  0.005,  ax2  is  large  and  bx^  is  small. 
Thus  the  larger  quantity  may  be  cut  back  to  the  number  of  significant 
figures  of  the  smaller. 

Therefore  find  21700000  x  1.53,  as  close  enough. 

2.  What  is  the  product  of  372  and  0.0001,  these  numbers  representing 
measurements  ? 

Here  the  first  number  shows  three  significant  figures  and  the  second 
only  one. 

Therefore  take  400  x  0.0001. 

The  rejection  error,  as  a  per  cent,  is  in  the  first  case  1|ft(7|T  +  i)  an(l 
in  the  second  case  i%°-  (?£7  +  £).     The  difference  is  slight. 


52  PLANE   TRIGONOMETRY.  [§34 

3.  What  is  the  relative  accuracy  of  measurement  in  each  factor  of  the 
products 

1.000  x  0.0001 ;  10.00  x  0.01 ;  1.000  x  0.0001  ? 

Find  the  per  cent  rejection  error  in  each  product. 

§  34.    Some  Extreme  Cases  and  Computation  Rules  based  on  them. 

(1)  Suppose  there  are  20  factors  in  a  product,  each  factor 
with  four  significant  figures,  and  each  factor  about  1000. 

The  accumulation  rejection  error  as  a  per  cent  is  about 

*t*  (tuW  +  ToV o  +  etc->  to  twenty  terms) 
or  lf>. 

(2)  Taking  now  the  other  extreme  case  of  four  significant 
figures,  where  each  factor  is  about  9999  (without  regard  to 
decimal  place). 

The  per  cent  rejection  error  is  now 

-^(9^9  +  W99  +  etc->  t0  twenty  terms) 
or  about  -^  of  lf>. 

Similarly,  for  numbers  of  five  significant  figures  the  results 
are  from  -^  of  1  f>  to  j^  of  1  fo,  and  so  on. 

From  these  results  arise  the  following  computation  rules 
occasionally  advised  : 

For  an  inaccuracy  ranging  from  -^  of  1  fo  to  If  in  the  final 
result,  use  four  significant  figures  in  each  factor  of  the  product 
and  in  each  partial  product  as  formed  and  in  the  final  product. 

When  the  range  of  inaccuracy  is  to  be  from  ^  of  1  fo  to  yj-g- 
of  1  fo,  use,  similarly,  five  significant  figures,  and  so  on. 

Since  a  product  with  20  terms  is  uncommon,  and  since  in 
the  majority  of  cases  the  number  of  factors  in  the  product  to 
be  found  is  quite  few,  the  rough  approximation  rules  given 
in  §  28,  for  the  determination  of  the  per  cent  error  in  the 
final  product,  will,  in  any  individual  case,  generally  shorten 
the  work  sufficiently  to  justify  finding  the  maximum  error 
likely  to  arise.  Then  the  number  of  significant  figures  to 
take  to  keep  within  the  prescribed  error  can  be  determined. 
Illustrative  examples  have  already  been  given  in  §  32. 


§35]  CALCULATION  VICES  AND   DEVICES.  53 

The  foregoing  computation  rules  are  sometimes  stated 
incorrectly,  thus :  "  For  an  inaccuracy  of  1  ^  or  worse,  use 
four  significant  figures,  etc."  The  error  cannot  be  worse 
than  1  f>  for  less  than  20  factors.  That  is  the  extreme  case, 
as  already  shown. 

§  35.    On  what  Forms  of  Products  Accumulation  Rejection  Error 
has  Least  Effect. 

We  have  already  shown  that  in  a  product  of  two  factors 
the  accumulation  rejection  error  is  least,  for  a  given  product, 
when  the  two  factors  are  the  same,  decimal  point  being  dis- 
regarded.    (See  §  29.) 

Similarly  for  three  factors  in  a  given  product,  the  effect 
of  such  error  is  least  when  A  =  B  =  (7,  and  for  four  factors, 
making  a  given  product,  when  A=  B—C=  D,  and  so  on. 

EXERCISES. 

1.  A  square  and  rectangle  of  about  the  same  area  have  their  sides 
measured  and  areas  calculated.  For  the  square,  the  side  is  24.82.  For 
the  rectangle,  the  sides  are  12.41,  49.64.  Determine  by  the  shortened 
process  of  multiplication  the  areas  of  each  and  the  per  cents  of  rejection 
error  in  each  area. 

2.  Consider  in  the  same  way  the  cube  whose  edge  is  12.36  and  the 
parallelopipedon  whose  edges  are  6.06,  4.09,  73.2. 

3.  Consider  in  the  same  way  two  cylinders  of  about  1450  cubic  inches 
volume,  where  in  one  r  =  6.043,  h  =  12.65,  and  in  the  other  r  =  3.022, 
h  =  50.60. 

4.  What  shaped  cylinder  can  be  measured  most  accurately  to  five 
places  for  volume  ?  Ans.    r  =  h  =  ir  =  3.1416. 

5.  What  circle?  Ans.     r  =  tt. 

6.  Which  can  be  measured  with  greater  accuracy  for  area,  a  square 
or  rectangle,  both  being  of  about  the  same  area  ? 

7.  For  volume,  cube  or  parallelopipedon,  both  being  of  about  the 
same  volume? 

8.  What  shaped  triangle  for  area,  the  base  and  altitude  being 
measured  ? 

9.  What  shaped  right-angled  triangle  for  area,  if  the  two  legs  are 
measured  ? 

10.    What  shaped  cone  ? 


54  PLANE   TRIGONOMETRY.  [§36 

8  36.    Approximate  Values  of and when  x  is  Small. 

3  vv  1  +  jrl-jr 

By  actual  division, =  1  -  x  +  x2  —  xz  -f-  Rv 

1  +  x 

By  actual  division, =  l+x  +  x2  +  x?  +  iL,  where  B 

1  —  x 

denotes  "  the  rest "  after  the  terms  written  down. 

If  x  is  less  than  unity,  x2,  Xs,  etc.,  are  smaller  than  x.  If 
x  is  0.0001,  x2  is  0.00000001,  and  so  on.  When  x  is  on  the 
verge  of  the  smallness  taken  into  account  in  any  measurement, 
x2,  x3,  etc.,  may  be  neglected. 

Thus,  approximated,     = =  1  —  #, 

1  +  x 

and =  1  +  x. 

1  —  x 

EXERCISES. 

1.  —  =  — - —  =  1.1,  nearly. 
0.9      1-0.1  J 

2.  -i-  = 1 =  1.01,  nearly. 

0.99      1-0.01  '  J 

3.  —?—  = 1 =  1.001,  nearly. 

0.999      1  -  0.001  J 

4/  — - —  =  1  -  0.0001  =  0.9999. 
1.0001 

§  37.    Approximate  Values  of  when  x  is  Small. 


l  l      ^ir    x\_i    2 

a  ±  x        f^x\a\        a)     a     a 


EXERCISES. 

i.  JL-l-JL. 

9.1      9      810 

2.  -J—  =  -1 ?—  =  0.0997,  nearly. 

10.03     10      10000  '  y 

3.  -A-  = 1 =  A  +  — 1 —  =  0.10001. 

9.999     10  -  0.001     10     100000 


§40]  CALCULATION   VICES   AND   DEVICES.  55 

§  38.    If  a  is  liable  to  an  error  x,  where  x  is  small,  to  what 

error  is  -  liable  ? 
a 

By  §  37  the  error  is  — . 
a2 

§  39.    If  a  is  liable  to  an  error  xx  and  b  to  an  error  xv  to 

what  error  is  the  quotient  j  liable  ? 

1  o 


?-•©■ 


This  is  a  product,  and  the  error  in  such  a  product  is 

a  times  the  error  in  -  plus  -  times  the  error  in  a. 
o  b 

^2      1         =  ax,,  +  bxx 

b*     b      l  W- 

Thus  the  error  in  the  quotient  -  is  that  in  the  product  ab 
divided  by  b2. 

As  a  per  cent  this  error  is — \ —,  which  is  the 

ab 

same  as  it  would  be  in  the  product  ab. 

Thus  all  that  has  been  said  concerning  products  hereto- 
fore holds  also  for  quotients,  so  far  as  per  cent  errors  is 
concerned. 

§  40.     The  Shortened  Process  of  Division. 
Special  Example  :  find  the  value  of  *      • 

The  first  step  is  to  determine  to  how  many  places  the 
division  may  be  carried  accurately,  assuming  the  terms  as 
representing  approximate  numbers  and  thus  subject  to 
rejection  error.  This  can  be  determined  by  calculating  the 
error  in  the  quotient  by  the  formula  of  §  39.  This  is  too 
much  trouble --as  much  as  performing  the  division  itself. 
It  is  better  to  get  the  rejection  error  approximately  as  a  per 
cent,  and  calculate  roughly  the  quotient,  and  thus  calculate 
the  inaccurate  place  in  the  quotient. 


+ 


fe) 


56  PLANE  TRIGONOMETRY.  [§40 

In  the  quotient  above,  the  per  cent  error  is  (by  §  27) 

100/      1 
2  U57853  ' 

or,  approximately,  -  •  —  =  zrrj  °f   1  °lo .      The   quotient  is 

roughly  2,  of  which  — —  of  1%  will  be  a  figure  in  the  fourth 

decimal  place.     The  division  may  thus  be  carried  accurately 
not  beyond  the  third  decimal  place. 

2.198 
7182)15785.3 
14364 
14213 
7182 
7031 
6464 


567 
574 

Explanation. 

(1)  Make  the  denominator  an  integer  by  multiplying  both 
numerator  and  denominator  by  an  appropriate  power  of  10. 
The  quotient  is  set  in  the  top  line,  and  so  that  the  decimal 
points  fall  the  one  under  the  other. 

(2)  In  carrying  out  the  division,  one  proceeds  as  in 
arithmetic  until  the  final  significant  figure  of  the  dividend 
is  used.  This,  above,  is  until  the  remainder  7031  is 
reached. 

(3)  When  such  figure  is  reached,  then,  instead  of  drawing 
down  zeros  in  the  dividend,  as  in  arithmetic,  the  divisor  is 
cut  one  figure,  making  it  718,  and  718  goes  into  7031,  9 
times,  giving  the  third  figure  of  the  quotient.  One  carries 
from  the  rejected  2  of  the  divisor  in  multiplying  718  by  9 ; 
9  times  2  is  18,  and  2  is  carried  to  the  9  times  8,  making 


§40]  CALCULATION   VICES   AND   DEVICES.  57 

74.  At  the  next  step  the  divisor  is  cut  to  71,  which  goes 
into  568,  8  times.  In  multiplying  by  8,  6  is  carried  from  8 
times  the  rejected  8  of  the  divisor. 

A  second  example  :    

F    673.2 

23.29 

6732)156780 

13464 

2214 

2020 

194 

135 

59 

60 


EXERCISES. 

Determine  to  how  many  places  the  divisions  following  should  be 
carried,  and  then  do  the  work  as  in  the  preceding  examples : 

15374.      327.3  .     427.    4J5 
624.2'    31.8384'    61.7'    7.3* 

The  teacher  may  assign  other  examples  at  pleasure. 

In  the  divisions  which  occur  in  practical  work,  arising  from 
measurements  intelligently  taken,  both  the  divisor  and  divi- 
dend will,  as  already  pointed  out,  show  the  same  number  of 
significant  figures,  and  the  limit  of  accuracy  takes  care  of 
itself  in  the  actual  division  by  the  disappearance  of  the 
divisor  itself  under  the  pruning  process  to  which  the  method 
subjects  it. 

A  number  of  examples  should  be  selected,  in  which  numera- 
tor and  denominator  show  the  same  number  of  significant 
figures,  the  number  of  places  to  which  the  division  may  be 
carried  accurately  determined  by  the  methods  given,  and 
then  the  division  by  the  shortened  process  should  be  carried 
out,  noting  that  the  divisor  vanishes  opportunely. 

Test  results  by  using  logarithms. 


58  PLANE   TRIGONOMETRY.  [§41 

§  41.  If  two  numbers  subject  to  rejection  error  are  multi- 
plied together  by  the  use  of  logarithms,  what  effect  have  such 
errors  on  the  logarithms  and  on  the  final  result  as  calculated 
by  logarithms  ? 

It  has  been  pointed  out  in  §  12  that  if  two  numbers  differ 
by  e,  a  small  number,  their  logarithms  will  differ  by 

(0.4342.  .  .).£". 

x 

Also  log  (abc  .  .  .  V)  =  log  a  -f-  log  &  +  •••+  log  I. 
Thus,  if  a,  b,  c'.  .  .  Z,  have  the  errors  xv  x2  .  .  .  #n,  the 
error  in  the  \og(abc  .  .  .  I)  is 


(0.4342.  .  .-j/a  +  Sl.  •  .3i\ 


In  the  case  where  the  #'s  are  rejection  errors,  this  becomes, 
as  in  the  cases  already  considered, 

(0.4342  .  .  .)/!       1      1  .         .   1_\ 
2  U     -B     C7-         xy 

Corresponding  to  this  error  in  the  log  of  the  product  will  be 
the  relative  error  in  the  product  itself, 


2\A     B     0  B. 


& 


This  is  exactly  the  same  as  would  be  the  relative  error 
due  to  rejection  error  if  the  multiplications  were  carried  out 
directly.     (See  §  25  and  §  31.) 

Consequently  the  use  of  logarithms  has  no  effect  on  the 
final  result  other  than  that  arising  from  rejection  error  in 
the  logs  themselves,  and  which  has  been  considered  already 
in  §  22. 

§  42.   Approximate  Values  of  Powers  and  Roots. 
When  x  is  small  compared  with  <z,  then 

(a  +  a;)*  =  a  +  2  ax ;     (a  +  x)*  =  a?  +  \  •  —x . 

2    a? 


§42]  CALCULATION  VICES  AND   DEVICES.  59 

(a  +  xf  =  a  +  3  ax ;     (a  4-  a:)*  =  at  +  \  —  • 

8    a' 
I        i       -> 

fa  +  x)n  =  a  +  nx;'  (a  +  x)n  =  an  +  -•  -—• 

a  ra 

If  a:  is  a  percentage  =  — —  (z  per  cent),  the  above  results 
become 

2  "      I  I         1         -1        z 

(a  +  a;)"  =  an  +  rcan  •  j^j ;     (a  +  a;)"  =  an  +  -  •  an  •  — -. 

Tims  the  per  cent  of  error  in  the  square  of  a  number  is, 
approximately,  twice  that  in  the  number;  in  a  cube,  three 
times  that  in  the  number,  and  so  on. 

The  per  cent  of  error  in  a  square  root  is,  approximately, 
one-half  that  in  the  number ;  in  a  cube  root,  one-third,  and 
so  on. 

EXERCISES. 


1.  If  x  =  0.000008168,  what  is  the  value  of  Vl  -  a;? 

Vl  -  x  =  1  -  I  x,  nearly, 
=  1  -  0.000004234, 
=  0.9999958.' 

2.  If  the  side  of  a  square  is  1.0002,  what  is  the  area  approximately  V 

(1.0002)2  =  1  +  2  x  0.0002  =  1.0004,  nearly. 

3.  Find  the  approximate  yalue  of  (1.0001)8. 

4.  Find  the  approximate  value  of  Vl.0003. 

5.  Find  the  approximate  value  of  V0.9999996. 

6.  If  the  mantissa  is  zero,  what  modification  is  made  in  the  rule  on 
page  11  for  getting  the  cologarithm?  What  is  the  colog  0.01?  Colog 
no)  ioooooo  7 


CHAPTER   III. 

ANGLES   AND   ANGLE-UNITS. 

§  43.    Angles  in  Formation  and  in  Sign. 

The  student  beginning  the  study  of  trigonometry  is  already 
familiar  with  the  word  angle.     Some  extension  of  his  ideas,  as 

to  size  and  sign  of  angles,  is 
essential.  Imagine  OA,  OP, 
to  be  two  straight  lines  hinged 
together  at  0  ;  and  that  OA  is 
held  fixed  in  position  while 
OP  turns  in  a  plane  about  0, 
starting  from  coincidence  with 
OA.  Then  the  amount  that 
OP  has  turned,  as  compared 
with  some  unit-turn,  is  a  meas- 
ure of  the  angle  between  OA 
and  OP.  OP  may  turn  as 
do  the  hands  of  a  clock  (clockwise),  or  in  the  opposite  way 
(counter-clockwise).  To  distinguish  between  the  two,  when 
this  is  desired,  angles  described  by  a  counter-clockwise  turn 
of  OP,  as  M,  are  called  positive  angles  ;  the  reverse,  like  N, 
negative  angles.  The  line  OA  is  called  the  initial  line  ;  OP, 
the  terminal  line,  or  simply  the  terminal,  of  the  angle.  When, 
therefore,  significance  is  attached  to  the  sign  of  an  angle, 
a  distinction  is  made  between  its  border  lines.  One  is  the 
initial  line ;  the  other,  the  terminal  line.  The  latter  must, 
for  positive  angles,  lie  in  the  order  of  a  counter-clockwise 
turn  from  the  former ;  the  reverse,  for  negative  angles. 


h 

/ 

/ 

V^       N 

1 
1 
1 

1 

i 

1            0 

Initial  \  line 

\ 

/ 

\ 

/ 

\ 

/ 

s 

/ 

Ns 

Fig.  1. 


§  44.    Adding  and  Subtracting  Angles. 

Angles  are  added  as  are  numbers.     To  add  A  to  B,  pictori- 
ally,  first  lay  out  A,  paying  attention  to  both  magnitude  and 

-    60 


§45]  ANGLES  AND  ANGLE-UNITS.  61 

sign.  Then  from  the  terminal  of  A  as  a  new  initial  line,  lay- 
out an  angle  equal  to  B  and  of  the  same  sign.  The  old 
initial  line  of  A  and  the  new  terminal  of  B  form  the  initial 
line  and  terminal  of  A  +  B.  This  applies  to  ±  A  +  (  -f-  i?) 
as  well  as  to  ±  A  +  ( —  i?) .  To  subtract  the  angle  B  from 
the  angle  A,  lay  from  the  terminal  of  A  an  angle  equal  in 
magnitude  to  B,  but  of  opposite  sign.  The  initial  line  of  A 
and  the  new  terminal  of  B  form  the  border  lines  in  order  for 
A  -  B.  This  applies  to  ±  A  -  (  +  B)  and  to  ±  A  ~(-B). 
In  Fig.  1,  OA,  OB  border  many  different  angles,  so  that  the 
expression,  angle  AOD,  is  indefinite.  However,  any  angle 
A  OB  is  the  sum  of  two  represented  by  A  OP,  P  OB.  It  is 
also  the  difference  of  two  represented  by  A  OP  and  BOP. 
So  is  AOP  equal  to  AOB  +  BOP,  or,  it  is  AOB-POB. 

EXERCISES. 

1.  What  is  the  method  used  in  geometry  for  constructing  an  angle 
equal  to  a  given  angle  ? 

2.  Draw  with  a  straight-edge  some  angles  at  random.  Construct  an 
angle  equal  to  the  sum  of  two  selected  positive  angles ;  then  to  three. 
Add  a  positive  and  a  negative  angle,  selected  at  will.  Subtract  a  positive 
angle  from  a  given  larger  positive  angle ;  from  a  given  smaller  positive 
angle ;  a  positive  from  a  negative  angle.  Make  the  number  and  variety 
of  exercises  sufficient  unto  proficiency,  but  not  unto  weariness. 

Note.  —  The  teacher  using  this  book  is  advised  against  assigning 
to  tiresomeness  exercises  that  are  alike.  Nothing  deadens  intellect  more 
than  having  to  do  too  much  of  a  tedious  thing,  which  one  can  see  readily 
how  to  do,  but  does  not  wish  to  do. 

* 

§  45.    The  Initial  Line  Par  Excellence  and  the  Quadrants. 

When  angles  are  considered  singly,  with  reference  to  size 
and  sign,  and  with  reference  to  certain 
related  magnitudes,  called  sines,  cosines, 
etc.,  which  vary  with  the  size  and  sign 
of  angles,  it  is  customary  to  take  as  initial 
line  a  right-hand  horizontal  line,  as  OA  in 
Fig.  1  or  Fig.  2.  The  point  0  is  called 
the  origin;  also  the  pole.     Continuing  OA  Fig.  2. 


II 

I 

0 

III 

IV 

62 


PLANE   TRIGONOMETRY. 


[§45 


backward  through  the  origin,  0,  and  drawing  through  0,  at 
right  angles  to  OA,  another  line,  and  extending  both  lines 
in  imagination  infinitely  in  both  directions,  we  divide  the 
plane  of  angles  into  four  quadrants.  These  quadrants  are 
numbered  and  named,  as  indicated  in  Fig.  2,  in  counter- 
clockwise order,  the  first  quadrant  (I),  the  second  quadrant 
(II),  the  third  quadrant  (III),   the  fourth  quadrant  (IV). 


> 

^ 

Angle  of  lei 
Fic 

,  quadrant 

k3. 

Angle 

Fi( 

of  2nd 

j.  4. 

Angle  of  3rd 

Fig.  5. 

/  Angle  of  4th 

Fig.  6. 

An  angle  whose  terminal  is  in  the  first  quadrant  (I),  no  mat- 
ter what  its  size  or  sign,  is  said  to  be  an  angle  of  the  first 
quadrant.  When  the  terminal  is  in  the  second  quadrant, 
the  angle  is  said  to  be  an  angle  of  the  second  quadrant ; 
similarly,  with  reference  to  the  third  and  fourth  quad- 
rants. In  all  these  cases  the  right-hand  horizontal  line  is 
reckoned  the  initial  line. 


/ 


M 


§  46.    Angles  unlimited  in  Size. 

In  geometry  the  angles  considered  are,  for  the  most  part, 

less  than  two  right  angles. 
In  trigonometry  no  limit  is 
set  upon  the  size  of  angles. 
That  is,  no  limit  is  set  upon 
the  extent  of  turn  of  OP  in 
*  Fig.  7.  It  may  pass  any  posi- 
tion any  number  of  times,  finite 
or  infinite,  and  in  either  direc- 
tion. In  Fig.  7  the  arrow  M 
indicates  the  smallest  positive 
Fig.  7.  angle,  M,  corresponding  to  that 


/ 


§47  a]  ANGLES  AND  ANGLE-UNITS.  63 

position  of  the  terminal.  If  T  denotes  one  complete  turn  of 
OP,  from  the  position  OA,  back  to  that  position  again,  then 
M±  T,  M±  2  T,  M±  3  T,  M±  nT  (where  n  is  any  integer), 
are  angles  whose  terminals  are  coincident  with  that  of  M. 
Denoting  by  the  arrow  N,  in  Fig.  7,  the  smallest  negative 
angle,  numerically,  corresponding  to  the  given  position  of 
the  terminal  of  M,  then  also,  evidently,  N±  T,  N±  2  T, 
N±nT,  are  angles  whose  terminals  are  coincident  with 
that  of  M. 

Thus  M+nT  is  a  general  formula  for  all  angles  whose 
terminals  are  coincident  with  that  of  M,  n  being  any  positive 
or  negative  integer. 

The  smallest  angle,  numerically,  locating  any  given  ter- 
minal, is  called  the  principal  angle  of  the  terminal. 


§  47.    Measuring  Angles. 

To  measure  an  angle  is  to  determine  its  relation  in  magni- 
tude to  some  unit-angle,  that  is,  its  number  (or  its  ratio  to  the 
unit-angle).  This  is  done  in  practice,  in  outdoor  work,  by 
means  of  graduated  circles,  with  some  means  of  "  pointing," 
more  or  less  accurate,  ranging  in  accuracy  from  the  u  sights  " 
on  two  uprights,  as  in  the  surveyor's  compass,  to  the  telescope 
of  a  transit  instrument.  Detailed  descriptions  of  such  appa- 
ratus will  be  found  in  books  on  surveying.  Angles  on  paper 
are  measured  by  means  of  a  protractor.  This  is  taken  up 
in  the  following  chapter.  We  are  concerned  here,  for  the 
present,  with  the  unit-angle. 

(a)  The  Sexagesimal  Angle  Measure,  or  Degree  Measure. 
In  America,  England,  and  Germany  (and  some  other  coun- 
tries) the  unit-angle  is  one  degree  (1°),  or  ^^  of  a  complete 
turn.  Fractional  parts  of  this  unit  are  expressed,  ordinarily, 
in  minutes  and  seconds  ;  sometimes  in  decimal  parts  of  a 
degree.  The  student  is  assumed  to  be  familiar,  from  his 
work  in  arithmetic,  with  the  notation  for  degrees,  minutes, 
and  seconds,  as  in  24°  13'  43",  and  with  the  table: 


64  PLANE   TRIGONOMETRY.  [§47  a 

60  seconds  make  a  minute, 
60  minutes  make  a  degree, 

as  also  with  the  conversions  which  may  arise  in  connection 
with  the  table. 

From  the  prominence  of  60  in  the  table,  this  system  of 
angle  measurement  is  frequently  called  the  Sexagesimal  Sys- 
tem ;  more  frequently,  however,  merely  the  Degree  Measure. 

To  the  Babylonians  is  ascribed  the  credit  of  originating  this  system. 
It  is  undoubtedly  of  very  great  antiquity.  The  "360  degrees  in  a  cir- 
cumference" dates  back  to  that  remote  period  when  astronomy  was  in  its 
infancy  and  men  thought  the  year  contained  360  days.  Thus  the  unit- 
angle,  one  degree,  is  but  what  was  thought  then  the  daily  step  (gradus) 
of  the  sun  in  his  apparent  annual  walk  around  the  ecliptic,  his  step  being 
each  day  about  twice  his  angular  diameter. 

Why  "  60  minutes  make  a  degree  "  and  "  60  seconds  make  a  minute," 
is  lost  in  obscurity.  A  fairly  good  guess,  but  only  a  guess,  is  that  the 
Babylonians  had  noticed  that  the  radius  of  a  circle,  used  as  a  chord, 
would  go  just  six  times  around  the  circle,  subtending  60  degrees,  thus 
making  60  a  sort  of  "  charmed  number." 

(5)  The  Centesimal  Angle  Measure,  or  the  Grade  System, 
or  French  System.  In  this  system  of  units,  the  right  angle 
is  divided  into  100  equal  parts,  instead  of  90,  as  in  degree 
measure,  and  one  hundredth  of  a  right  angle  is  the  primary 
unit-angle.  This  unit  is  called  a  grade.  A  minute  and  a 
second  in  this  system  are,  respectively,  the  hundredth  and 
ten-thousandth  of  the  grade,  or, 

100  seconds  make  a  minute, 
100  minutes  make  a  grade. 

In  this  system  the  angle  is,  therefore,  always  expressed 
decimally.  The  expression  19g.3552  is  19  grades,  35  min- 
utes, 52  seconds,  while  3g.0407  is  3  grades,  4  minutes,  7 
seconds.  Also  12g.3456  is  1234'. 56  and  123456",  the  unit 
being  changed  by  merely  moving  the  decimal  point.  The 
decimal  system  has  here,  as  in  all  other  cases,  great  advan- 
tages in  adaptability  to  calculations. 

Can  you  mention  any  serious  practical  difficulty  which 


^sjl 


/ 


§47c]  ANGLES  AND  ANGLE-UNITS.  65 

stands  in  the  way  of  a  change  from  degree  measure  to  grade 
measure  in  America? 

(<?)  Radian  Measure,  or  Circular  Measure,  and  ir-Measure. 
(1)  While  the  degree  and  grade  are  the  units  for  practical 
mathematics,  as  in  surveying  and  astronomy,  in  theoretical 
mathematics,  for  a  reason  which  the  student  is  not  yet  in  a 
position  to  appreciate  (it  will  appear  later),  another  unit, 
called  the  radian,  is  used.  The  radian  is  the  angle  sub- 
tended at  the  centre  of  any  circle  by  an  arc  of  its  circumfer- 
ence equal  in  length  to  its  radius.  Since  in  unequal  circles 
the  arcs  subtending  equal  central  angles  are  to  each  other 
as  the  radii  of  the  circles,  the  radian  is  the  same  for  all 
circles.  Later  (§  49)  it  will 
be  shown  that  3.14  •••,  or 
7r-radians,  make  180°.  Fre- 
quently 7r-radians  are  taken        /  ^ ~Q 

as  the  unit.     This  measure      / 
might  be  called  the  7r-meas-     /  / 

ure.  |  l 

(2)  Adjoining    is   a  pic-     \         \ 
ture   of    the   radian    as    an       \ 
angle,  the  angle  A  OB,  where         \ 

arc  AB  =  radius  OA 

,       .  Angle  AOB=5n°n'U.$' 

and  where  =Radiaa 

arc  A'B'  =  radius  OA' .  Fig.  8. 


LABORATORY  EXERCISES. 

Construct  five  circles  of  pasteboard,  and  of  different  sizes,  with  smooth 
edges.  Wrap  on  each  circumference  a  string  equal  in  length  to  its 
radius,  and  cut  out  from  each  circle  the  corresponding  sector.  See  if 
these  angles  fit  each  other.  Get  the  size  of  a  radian  thus  fixed  in  the 
mind.     Measure  this  angle  with  the  protractor. 

Devise  a  plan  for  dividing  a  radian  into  tenths.  Use  the  radian  you 
have  constructed  to  get  in  radian  measure  the  values  of  30°,  45°,  60°,  90°, 
120°,  135°,  150°,  180°,  to  the  nearest  tenth.  What  relation  have  your 
answers  to  tt  =  3.14?  Determine,  mechanically,  how  many  radians  make 
360°.     How  are  rims  and  rods  graduated  ? 


66 


PLANE   TRIGONOMETRY. 


[§47 


(3)  The  Radian  Protractor  (Polar  Coordinate  Paper).  In 
Fig.  9  we  give  a  central  clipping  from  the  radian  polar 
coordinate  paper  of  Professor  B.  F.  Groat  of  the  University 
of  Minnesota.  The  divisions  are  in  radians  and  tenths  of  a 
radian.*  • 


0  and  6.28 


Fig.  9. 


EXERCISE. 


Set  arrows  on  the  diagram  at  45°,  60°,  90°,  120°,  135°,  150°,  180°,  210°, 
225°,  240°,  270°,  300°,  315°,  330°,  360°,  and  estimate  with  the  eye  the  cor- 
responding values  in  radians,  tenths,  and  hundredths.  Compare  these 
values  with  the  calculated  values  obtained  by  §  48. 


*  This  paper  is  recommended  for  use,  with  this  book,  in  drawing  diagrams 
to  scale  (Chap.  IV.),  locating  points  in  polar  coordinates  (§  55  ?>),  curve- 
tracing  (Graphs,  §§  102,  132),  etc.  Here  it  should  be  used  to  estimate  with 
the  eye  the  size  in  radians  of  degree-angles,  that  the  student  may  really  know 
what  a  radian  is.  Groat's  Polar  Coordinate  Paper  can  be  supplied  by  the 
H.  W.  Wilson  Co.,  Minneapolis,  Minn.,  in  both  radian  measure  and  degree 
measure. 


§47c]  ANGLES  AND   ANGLE-UNITS.  67 

(4)  The  radian  measure  of  an  angle  can  also  be  pictured  as 
a  line  (arc).     Let  A  OP  be  the  given 

angle.    Draw  about  0  the  unit  circle  : 

the  portion  of  this  circle-arc  included 

between  the  initial  line  and  terminal 

line  of  the  angle  is  a  picture  of  the 

radian  measure  of  the  angle ;  that  is, 

the  number  representing  the  length  of 

the  arc  BD  (between  the  arrows)  is  the 

same  as  that   representing   the  angle  FlG  10 

BOD.     For 

arc  BD         angle  BOD      n^n  •        j- 

— - — : =  — ^— — =  BOD  in  radian  measure. 

radian  s  arc  radian 

But  the  radian's  arc  is  the  radius,  which  is  here  unity. 

•\  arc  BD—  radian  measure  of  the  angle  BOB. 

Since  the  scale-unit  is  quite  arbitrary,  so  is  the  size  of  the 
adjoining  picture. 

LABORATORY  EXERCISE. 

Construct  a  pasteboard  circle  of  one  inch  radius  (or  of  one  foot  radius). 
Lay  out  five  central  angles  on  the  circle.  Cut  strings  equal  in  length  to 
arcs  of  these  angles.  Get  the  lengths  of  the  strings  ou  the  inch  rule  (or 
in  decimals  of  a  foot,  if  a  circle  of  one  foot  radius  is  used).  These 
lengths  are  the  circular  measures  of  the  angles.  Measure  the  angles 
with  the  protractor.  Reduce  the  protractor  measure  to  circular  measure 
(§  48).     Compare  results  with  string  lengths. 

(5)  Similar  to  the  preceding  is  the  surface  picture  of  the 
spherical  measure  of  a  solid  angle.  Draw  about  the  vertex 
of  the  solid  angle  a  sphere  of  unit  radius  :  the  portion  of  the 
surface  of  this  sphere  included  within  the  solid  angle,  and 
bordered  by  arcs  of  great  circles,  is  the  spherical  measure  of 
the  solid  angle. 

(6)  It  is  becoming  customary,  as  is  done  in  this  book,  to  denote 
angles  expressed  in  radian  (circular)  measure  by  the  small  letters  of  the 
Greek  alphabet,  while  the  capital  letters  of  the  English  alphabet  are  used 
to  denote  angles  expressed  in  degree  measure.     Since  many  students 


68 


PLANE   TRIGONOMETRY. 


[§47c 


taking  up  trigonometry  are  not  familiar  with  the  Greek  letters,  we  give 
here  the  Greek  alphabet  for  reference.  No  attempt  to  memorize  it  is 
recommended.  When  the  student  meets  a  new  Greek  letter  acquaint- 
ance (there  are  but  few  of  them  not  met  with  in  mathematical  journals 
and  astronomical  publications),  he  can  look  it  up  here. 


GREEK  ALPHABET. 


A 
B 

r 

A 
E 
Z 

H 
© 

I 

K 
A 
M 


a 

y 
8 

e 
t 
V 
0or 

i 

K 


alpha 

N 

beta 

S 

gamma 

O 

delta 

n 

epsilon 

p 

zeta 

5 

eta 

T 

theta 

Y 

iota 

<S> 

kappa 

X 

lambda 

* 

mu 

Q 

nu 

xi 

omikron 

pi 

rho 

sigma 

tau 

upsilon 

phi 

chi 

psi 

omega 


(7)   Circular  Ares  in  Radian  Measure. 

given  arc   on 


v 


Take  AB  as  any 

a   circle  of  radius 

OA  =  r.      Draw  the   unit    circle, 

CBO,  concentric  with  the   given 

circle.     Then 

nrc  AB  _  radius  OA  __  radius  OA 
arc  CD     radius  00  1 

,\   arc  AB  =  radius  x  arc  OB. 

But  arc  OB  is  the  circular 
(radian)  measure  of  the  angle  A  OB 
(as  shown  in  (4),  page  67)  or  6. 

.*.   arc  AB  =  r  -6. 

That  is,  the  are  of  any  circle  is  its  radius,  r,  multiplied  by  the 
circular  measure,  0,  of  the  central  angle  which  it  subtends. 

arc  AB 


Fig.  11. 


Also 


6  = 


that  is,  the  circular  measure  of  an  angle  can  be  expressed  as  the 
ratio  of  a  line  to  a  line  (the  radius),  making  it  of  the  same  char- 


§48  a]  ANGLES   AND   ANGLE-UNITS.  69 

acter  as  the  ratios  which  form  later  the  chief  study  of  this  booh. 
Herein  lies  the  importance  of  this  system  of  measurement 
in  theoretic  mathematics. 

LABORATORY  EXERCISE. 

Construct  from  pasteboard  a  circle  of  one  inch  radius.  Construct  five 
other  circles  of  pasteboard,  some  larger,  some  smaller,  than  the  inch 
circle.  Lay  out  equal  central  angles  on  all  the  circles.  Measure  with 
strings  the  lengths  of  the  corresponding  arcs.  Get  the  lengths  of  the 
strings  (in  inches)  and  the  lengths  of  the  radii.  Show  that  for  each  circle 
the  length  of  the  arc  is  its  radius  times  the  length  of  the  arc  on  the  inch 
circle. 

(8)  Similarly,  the  portion  of  the  surface  of  any  sphere, 
when  the  boundary  lines  are  great  circles,  is  r2  times  the 
spherical  measure  of  the  corresponding  solid  angle. 

(9)  Areas  of  Circular  Sectors  in  Terms  of  Radian  Measure 
of  their  Angles.  It  is  shown  in  geometry  that  the  area  of 
such  a  sector  is  one-half  its  radius  times  its  arc. 

.\  area  =  %r  •  r6=\  r2^, 

where  6  is  the  circular  measure  of  the  angle. 

(10)  Similarly,  the  volume  of  a  spherical  sector  is, 

J  r .  tty  =  i  rty 

where  <j>  is  the  spherical  measure  of  its  solid  angle. 


§  48.   Conversion  of  Angles  from  One  Measure  to  Another. 

(a)  Degrees  to  Grades  and  Vice  Versa.  One  right  angle  is 
90  degrees  and  is  100  grades.  Let  x  denote  the  number  of 
degrees  in  an  angle,  and  y  the  number  of  grades  in  the  same 
angle. 

.%£-¥•-*+.•!*  (1) 

tmdx=£jy  =  y-1\y.  (2) 

Thus,  by  (1),  to  convert  degrees  and  decimal  parts  of  a 
degree  to  grades,  add  one-ninth  of  the  given  number  to  itself. 


70  PLANE   TRIGONOMETRY.  [§  48  a 

When  minutes  and  seconds  are  given,  they  should  first  be 
changed  to  the  decimal  part  of  a  degree. 

And  by  (2),  to  convert  grades  to  degrees  and  the  decimal 
part  of  a  degree,  subtract  one-tenth  of  the  given  number  of 
grades  from  itself. 

This  is  not  a  matter  of  very  great  importance.  The  teacher  may 
assign  a  few  practice  examples,  selected  at  random. 

(5)  Degrees  to  Radians  and  Vice  Versa.  This  is  a  matter 
of  considerable  importance.  Denoting  by  it  the  number 
3.14159  •••,  as  in  geometry,  the  circumference  of  a  circle 
of  radius  r,  is  2  7rr,  and  any  two  arcs  of  the  same  circle 
are  to  each  other  as  the  central  angles  which  they  subtend. 

radian  radius  r         1 


4  right  angles      circumference      2  irr      2tt 

__  4  right  angles  '  180°  _     180° 
...  radian-  g-  -  — -  3.14159  '"* 

.\  radian  =  57°. 2957  •••  degrees,  or,  57°. 3,  approximately, 
=  3437'.747- 
=  206264".806  -  =  57°  17'  44".8. 

.*.  The  rule  for  converting  degrees  to  radians : 

(i)  When  the  angle  is  in  degrees  and  decimals,  divide  by 
57.2957 •••,  cutting  according  to  the  accuracy  desired. 

(ii)  When  the  angle  is  expressed  in  minutes  and  decimals, 
divide  by  3437.747  •••,  cutting  according  to  the  accuracy 
desired. 

(iii)  When  it  is  given  in  seconds,  divide  by  206265. 

Division  by  the  preceding  numbers  is  the  same  as  multi- 
plication by  the  following,  respectively : 

(iv)  For  degrees,  0.017453 
(v)    For  minutes,  0.000291  -. 
(vi)  For  seconds,  0.00000485  -. 
When  the  relation  of  degree  measure  to  radians  is  merely 


§48  6]  ANGLES  AND  ANGLE-UNITS.  71 

to    be    indicated,    the   work    being    left    unperformed,    we 
write : 

r   XX  x'  = 


180    '  10800  648000 

To  change  radian  measure  to  degree  measure,  use  the  pre- 
ceding divisors  for  multipliers  and  the  multipliers  for  divisors. 

When  numbers  indicate  radians,  it  is  customary,  in  case 
special  attention  is  to  be  called  to  the  fact,  to  use  r  or  c  as 
an  index.     Thus  2r  or  2C  would  mean  two  radians,  or  114°.59. 

EXERCISES. 

1.  Express  in  degrees  (decimally)  the  angles :  3r ;  2r.5 ;  4r.7 ;  4r.23. 
Test  the  results  by  working  them  backwards. 

2.  Express  in  radians  the  angles:  180°;  360°;  90°;  45°;  30°;  60°; 
235°;  270°;  19F.37;  424°.76;  5°.39;  17°  46' 35";  5°  43' 26";  10';  10"; 
1°;  1';  1".     Test  by  working  backwards. 

3.  If  the  radius  of  a  circle  is  57.3  feet,  how  long  is  the  arc  whose 
central  angle  is  1°?  If  3437.7  feet,  how  long  for  1'?  If  206265  feet, 
how  long  for  1"  ? 

4.  If  a  man  6  feet  tall,  leaning  on  the  rim  of  a  circle,  covers  the  arc 
of  a  radian,  what  are  the  circumference  and  radius  of  the  circle  ? 

5.  If  an  object  subtending  an  angle  at  the  eye  of  2£'  is  the  smallest 
object  that  is  visible,  how  far  away  is  a  brick  building  when  the  hori- 
zontal lines  of  plaster  are  just  discernible? 

6.  Find  the  ratio  of  13°  24'  56"  to  4.57  radians. 

7.  Assuming  the  radius  of  the  earth  as  4000  miles,  find  the  length 
of  a  radian  on  the  equator ;  of  a  degree ;  of  1' ;  of  1". 

8.  The  circular  measure  of  two  of  the  angles  of  a  triangle  are  £  and 
§ ;  what  is  the  third  angle  in  degrees  and  in  radians  ? 

9.  The  angles  of  a  quadrilateral  are  in  arithmetic  progression,  and 
the  greatest  is  double  the  least;  express  each  angle  in  radians  and  the 
least  angle  in  degrees. 

10.  The  angles  of  a  triangle  are  in  arithmetic  progression,  and  the 
number  of  radians  in  the  least  angle  is  to  the  number  of  degrees  in 
the  mean  as  1 :  120 ;  what  are  the  angles  in  radians? 

11.  The  diameter  of  the  whispering  gallery  in  St.  Paul's  is  108  feet; 
how  long  is  the  arc  of  a  radian?  Of  ir  radians?  What  is  the  area  of 
the  circular  sector  whose  arc  is  a  radian  in  this  circle? 


72  PLANE   TRIGONOMETRY.  [§49 

12.  The  large  hand  of  the  Westminster  clock  is  11  feet  long ;  how- 
many  yards  does  its  moving  extremity  travel  in  passing  over  a  radian? 
What  area  is  described  ? 

13.  What  is  the  circumference  of  the  circle  on  which  the  radian  arc  is 
3  inches?  What  is  the  area  of  the  sector  of  this  circle  whose  arc  is  the 
radius? 

14.  The  apparent  diameter  of  the  moon  is  30'  and  that  of  the  sun 
is  32' ;  how  many  moons,  placed  tangentially  to  each  other  on  the  cir- 
cumference of  a  circle,  would  it  take  to  cover  a  radian?  How  many 
suns  ?  Look  in  the  almanac  for  the  time  of  sunrise  and  sunset  to-morrow, 
and  determine  how  many  suns,  placed  tangentially,  would  cover  the  day- 
arc  of  the  sun's  motion  that  day.     The  night-arc. 

15.  A  railway  train  is  going  at  the  rate  of  60  miles  an  hour  on  a 
circular  curve  whose  radius  is  f  of  a  mile  ;  how  long  does  it  take  to  pass 
over  a  radian  ? 

16.  How  long  does  it  take  the  minute  hand  of  a  clock  to  pass  over  a 
radian  ? 

17.  If  7r  is  taken  as  %?-,  what  common  fraction  expresses  the  number 
of  degrees  in  a  radian?   'What  if  it  is  taken  as  fff  ? 

18.  What  is  the  area  of  a  circular  sector  whose  angle  is  §r,  the  radius 
being  10  feet  ? 

19.  A  radian  being  206265",  at  what  distance  does  1  foot  (assumed  as 
a  circular  arc)  subtend  an  angle  of  1"  ?    Of  1'  ?    Of  1°  ? 

20.  The  circumference  of  a  circle  is  divided  into  5  parts  which  are  in 
arithmetic  progression,  and  the  greatest  part  is  6  times  the  least ;  find 
in  radians  the  angles  subtended  at  the  centre  by  the  parts. 

§  49.   Special  Angles  in  Radian  Measure  (ir-Measure). 

By  (5)  of  the  preceding  section,  180°  in  radian  measure  is 

— -  times  180  radians,  or  ir  radians. 

180 

Thus  the  ratio  of  180°  to  the  unit  of  radian  measure  is 
the  same  as  the  ratio  of  the  circumference  of  a  circle  to  its 
diameter,  or  3.14159  •  •  •.  Consequently  it  is  the  Greek  letter 
representative  (see  (6),  page  67)  of  the  angle  which  is  the 
equivalent  of  a  half-turn,  or  180°.  It  is  customary  to  write 
180°,  in  circular  measure,  simply  as  7r,  instead  of  7rc,  or  7rr. 
Thus: 


50] 


ANGLES  AND  ANGLE-UNITS. 


73 


360°  =  2  7T 
540°=  3  7T 
720°  =  4  7T 


90°  = 


IT 


45°=? 


30c 


6 


±  180°  =  ±  7T  ±  60°  =  ± 


IT 


120°  =  f7T 


270°  =  1 7T 


240°=  4  7T 

135°=  |tt,  etc. 


EXERCISES. 

1.  Express  in  degrees,  in  radians,  and  in  terms  of  w  radians,  the  central 
angles  subtended  by  the  sides  of  each  of  the  first  twelve  regular  polygons. 
Express  in  terms  of  7r  radians  the  angles  whose  terminals  are  bisectors 
and  trisectors  of  the  quadrants. 

2.  Do  the  same  for  external  angles  of  the  same  polygons. 

3.  What  is  the  central  angle  of  a  regular  polygon  of  n  sides  in  terms 
of  7r  ?    In  radians  ?    In  degrees  ? 

4.  Show  that  A  °  expressed  in  7r-measure  is  - —  it. 

v  180 

5.  Reduce  some  numerical  angles  in  degree  measure  to  ir-measure. 


LABORATORY  EXERCISE. 

Construct  of  pasteboard  a  circle  of  one  foot  radius.  Lay  out  the 
angles  noted  in  §  49.  Use  strings  to  measure  the  arcs,  get  the  lengths 
of  the  strings  (in  feet),  and  test  the  above  results  numerically,  in  com- 
parison with  7r  =  3.1. 

§  50.  Special  Terminals  located  in  Radian  Measnre  (ir-Measure). 

If  n  is  any  positive  or  negative 
integer  (including  zero),  the  angles 
2  nir  radians  (or,  as  it  is  usually  writ- 
ten, 2mr)  will  have  their  terminals 
coincident  with  the  initial  line  OA. 
Thus  if  a  is  any  angle  whose  terminal 
is  given,  all  the  angles  2  mr  +  a  will 
have  that  same  terminal.  The  lines 
of  the  adjoining  diagram  indicate  the 
quadrant  and  semi-quadrant  lines.     Thus  all  angles 


74  PLANE   TRIGONOMETRY.  [§  50 

with  terminal  OA  have  the  form  2»7*,  as  0,  2  7r,  4-7T,  etc.; 
with  terminal  Oi?,  the  form  (2n  -f  l)7r,  as  7r,  37r,  5  7r,  etc.; 

with  terminal  00,  the  form  (2w  +  J)7r,  or  (4w  +  1)^, 

aS2'-2-'^'etC-; 
with  terminal  OB,  the  form  (2  w  -f-  |)7r,  or  (4  w  -f-  3)^, 

u 

37T     7  7T     11  7T 

aS^"'"2"'^"'etC-; 
with  terminal  OE,  the  form  (2  w  -f  ^)7r,  or  (8  w  4- 1)  —, 

7T     9  7T     17  7T        . 

as  -,  — ,  — ,  etc.; 

with  terminal  O^7,  the  form  (2  n  +  |)7r,  or  (8  n  -f  3)  j, 

3tt  11  tt  19  tt      , 

asT1,"5^  ~T~' etc,; 

with  terminal  06r,  the  form  (2  w  +  |)7r,  or  (8  w  +  5)—, 

5  7T    13  7T    21  7T        , 

as  — ,  -^—,  -j-,  etc. ; 
with  terminal  OH,  the  form  (2  n  +  |)tt,  or  (8  n  -f-  7)—, 

7  7T    15  7T    23  7T        . 

as  — ,  — ,  — >  etc. 

EXERCISES. 

1.  Show  that  if  n  is  an  integer  running  from—  go  to  +  oo,  2  n  +  1  and 
2  n  —  1  represent  the  same  set  of  numbers.  Find  expressions  for  every 
alternate  even  number ;  every  alternate  odd  number ;  every  fourth  even 
number ;  every  fourth  odd  number. 

2.  Consider  the  first  twelve  regular  polygons  as  central  at  0,  with 
one  vertex  on  the  initial  line,  and,  denoting  by  the  symbol  (VI,  4)  the 
terminal  line  for  the  fourth  vertex  of  the  hexagon,  counting  that  on 
the  initial  line  as  first  and  the  order  being  counter-clockwise,  write  the 
general  expression  for  all  angles  whose  terminal  is  that  of  any  specified 
vertex  of  any  one  of  these  regular  figures. 

For  example,  for  (VI,  4)  it  is  (2  n  +  4  x  £)tt  or  (12  n  +  4)§  tt.- 


§50]  ANGLES   AND  ANGLE-UNITS.  75 

3.  Taking  four  different  angles  (two  positive  and  two  negative) 
whose  terminal  is  OE  in  Fig.  12,  write  out  from  the  general  formula 
2  mr+  a,  where  a  is  each  of  the  four  angles  in  turn,  four  sets  of  ten  angles 
each,  placing  them  in  parallel  vertical  columns,  with  the  angles  obtained 
by  the  same  value  of  n  in  the  same  horizontal  line.  Compare  the  sets  and 
see,  if  n  were  given  all  integer  values  from  positive  infinity  to  negative 
infinity,  whether  or  not  the  four  sets  of  angles  would  be  identical.  Can 
one  set  be  brought  to  coincidence  with  any  other  set  by  sliding  ?  Is  the 
proposition  true  that  if  a  is  any  angle  of  a  given  terminal,  the  formula 
2  rnr  +  a  will  give  all  the  angles  corresponding  to  this  terminal  and  only 
these  ? 

4.  State  for  each  of  the  following  angles  the  quadrant  to  which 
it  belongs : 

1*5      -jp      W;     |*i     9tt;      -|tt;     -|;      -|tt;      -f*J     n*"+?; 

(2n  +  l)7r+§7r;  (2n-l)7r-327r;  2iMr+|j  2«r-|j   -§*■;  -§7r;   _|» 

5.  State  for  each  of  the  following  angles  the  quadrant  to  which  it 
belongs:  50°;  120°;  240°;  396°;  -50°;  -60°;  -101°;  -380°; 
-1000°;  -1080°;  -90°;  90°;  180°;  -180°;  270°;  360°;  425°;  370°; 
425°;    590°;    -750°;    -39°. 

6.  Give  for  each  of  the  angles  of  Exs.  4  and  5,  two  other  angles  which 
have  the  same  terminal,  expressing  those  for  Ex.  4  in  radian  measure. 

7.  Give  the  general  expression  in  degrees  for  the  difference  of  all 
angles  whose  terminals  are  coincident ;  the  general  expression  in  circular 
measure. 

8.  Give  the  general  expression  in  degree  measure  for  all  angles  whose 
terminals  are  symmetric  to  the  vertical  with  the  terminal  of  30°;  the 
general  expression  in  circular  .measure. 

9.  Give  in  degree  measure  the  general  expression  for  all  angles  whose 
terminals  coincide  with  that  of  45° ;   —  45°. 

10.  Give  in  degree  measure  the  general  expressions  for  all  angles 
which  have  for  terminals  the  border-lines  of  the  quadrants ;  the  general 
expressions  in  circular  measure  in  terms  of  ir. 

11.  Give  in  degree  measure  all  angles  whose  terminals  are  coincident 
with  that  of  ^4°;  symmetric  to  that  of  A0,  with  the  vertical;  with  the 
horizontal. 

12.  Give  in  all  three  systems  of  angle  measure  two  angles  whose  ter- 
minals are  symmetric  to  the  initial  line ;  to  the  90°  line  ;  to  the  180°  line ; 
to  the  270°  line ;  to  the  lines  bisecting  the  quadrants. 


76  PLANE   TRIGONOMETRY.  [§50 

13.  Give  the  general  expressions  for  some  selected  three  of  the  angles 
of  Ex.  2. 

14.  What  are  the  expressions  for  the  length  of  a  circular  arc  and  for 
the  area  of  the  corresponding  sector?  Selecting  two  angles  involving 
degrees,  minutes,  and  seconds,  calculate  the  lengths  of  the  corresponding 
arc  and  areas  of  the  corresponding  sectors,  on  a  circle  of  10  inch  radius. 

15.  Find  the  length  of  the  arc  subtending  an  angle  of  2r  on  a  circle  of 
8  inch  radius.  Express  in  terms  of  radians  the  corresponding  lengths  and 
areas  for  30°,  45°,  60°,  120°,  135°,  180°,  210°. 

16.  If  the  radius  of  a  circle  is  100  feet,  what  angle  in  radians  does  a 
10  foot  arc  subtend?  Express  the  same  result  in  terms  of  ir  radians. 
Make  up  and  solve  four  other  examples  like  this. 

17.  Find  in  degree  measure,  grade  measure,  and  radian  measure  the 
angle  between  the  hour  and  minute  hands  of  a  clock  at  20  minutes  of  6 ; 
at  2 :  30. 

18.  The  angle  of  a  circular  sector  is  22 1°  and  the  diameter  of  the  circle 
is  10  feet ;  what  is  the  area  of  the  sector  and  the  length  of  its  arc? 

19.  The  area  of  a  sector  of  a  circle  of  unit  radius  is  10  square  feet; 
what  is  its  angle  in  all  three  measures  ? 

20.  A  strip  of  paper  a  mile  long  is  rolled  tightly  into  a  circular  cylin- 
der. The  paper  is  0.001  inch  thick ;  what  is  the  radius  of  the  cylinder 
and  the  volume  of  the  cylindrical  sector  whose  angle  is  a  radian  ? 

21.  Three  circles,  each  of  10  inch  radius,  are  tangent  to  each  other ;  find 
the  length  of  the  arcs  between  the  points  of  tangency,  the  area  of  the 
sectors  and  the  area  bounded  by  the  circular  arcs,  assuming  ir  —  -\2-. 

22.  If  a  geographical  mile  on  the  earth's  surface  subtends  1'  at  the 
centre  of  the  earth  (radius  3960  miles),  how  far  off  is  the  sun  if  the 
earth's  radius  subtends  at  the  centre  of  the  sun  an  angle  of  8.76"  ? 

23.  The  moon  subtends  an  angle  of  about  30'  from  the  centre  of  the 
earth,  and  is  about  60  earth's  radii  distant ;  what  is  its  diameter,  approxi- 
mately, and  what  angle  does  the  earth  subtend  at  the  moon  ? 

24.  The  radius  of  a  circle  is  8  feet;  how  many  radians  does  an  arc  of 
13.2  feet  subtend? 

25.  If  the  number  of  degrees  in  one  angle  of  a  triangle  equals  the 
number  of  grades  in  another  and  the  number  of  radians  in  the  third, 
what  is  that  number  ? 

26.  Show  that  mr  +  -  will  give  angles  whose  terminals  are  either  on 

the  upright  vertical  or  downright  vertical,  n  being  any  positive  or  nega- 
tive integer. 


CHAPTER  IV. 

CONSTRUCTION  OF  ANGLES  AND  OP  STRAIGHT-LINE 
DIAGRAMS  TO  SCALE,  AND  THE  MEASUREMENT  OF 
ANGLES. 

[Note  to  the  Teacher.  —  It  is  advised  that  in  solutions  of  examples 
diagrams  to  scale  be  made  and  the  ungiven  parts  be  measured  and  compared 
with  the  calculated  parts  as  a  test.  The  diagram  will  serve  as  a  check  on 
numerical  solutions.  Even  a  rough  free-hand  sketch  will  serve  often  as  a 
check  on  results,  giving  a  "  common  sense  "  check,  saving  one  from  writing 
139  for  1.39,  and  the  like.  Drawing  to  scale  is  valuable  exercise  in  itself, 
in  preparation  for  "graphic  solutions,"  so  useful  in  all  engineering  work.'] 

§  51.  The  Protractor  (in  Degree  Measure  and  in  Radian  Measure). 

The  protractor  is  an  instrument  for  laying  out  angles  of  a 
given  size  on  a  diagram  and  for  determining  the  size  of  the 
angles  of  a  given  diagram.* 

(a)  To  lay  out  at  a  Given  Point  on  a  Straight  Line  a  Griven 
Angle.  Place  the  straight  edge  of  the  protractor  on  the  given 
line  and  the  middle  point  of  the  straight  edge  at  the  angle- 
vertex.  Then  with  a  sharp  pencil,  or  with  a  pin,  make  a 
dot  opposite  the  given  angle  on  the  circular  rim  of  the 
protractor.  Connect  the  dot  and  the  angle-centre  by  a 
straight  line. 

(6)  How  to  measure  a  given  angle  with  the  protractor  will 
occur  to  the  student  without  direction. 


*  A  protractor  (in  degree  measure)  and  a  scale  of  equal  parts  in  inches  and 
fractions  of  an  inch  should  be  used  in  connection  with  this  book.  Groat's 
Coordinate  Paper  in  degree  measure  and  in  radian  measure  furnishes  a 
cheap  protractor,  and  can  be  used  to  fix  in  the  mind  the  ability  to  estimate 
angles  as  well  as  measure  them.  Protractors  can  be  bought  made  of  paper, 
of  horn,  of  brass,  etc.  See  Johnson's  "Surveying,"  or  the  catalogues  of  the 
instrument  makers,  as  that  of  Dietzgen,  New  York. 

77 


78  PLANE   TRIGONOMETRY.  [§52 


LABORATORY  EXERCISES. 

1.  Make  radial  clippings,  from  Groat's  Coordinate  Paper,  of  angles  of 
various  sizes,  and  study  them  until  you  feel  able  to  estimate  fairly  well, 
in  degree  measure,  radian  measure,  and  7r-measure,  the  size  of  any  angle 
selected  at  random. 

2.  Construct  some  angles  at  random.  Guess  at  their  size.  Write 
down  your  guess.  Measure  the  angles  and  test  yourself  as  a  guesser  on 
angles.     Repeat  the  process  until  you  feel  like  congratulating  yourself. 

3.  Construct  angles  of  special  sizes  until  you  can  readily  estimate  by 
eye  the  number  of  degrees  and  the  number  of  radians  in  a  given  angle. 

4.  Construct  the  angles  of  the  first  ten  regular  polygons,  beginning 
with  a  triangle.     Construct  a  radian  and  a  degree. 

5.  What  angle  is  subtended  at  the  end  of  a  pencil  line  of  average  width 
by  the  width  at  different  points  along  the  line  ? 

6.  What  angle  is  subtended  by  the  opposite  edges  of  a  chalk-spot 
just  visible  on  a  blackboard  ?  What  angle  is  subtended  by  the  distance 
between  the  pairs  of  a  double  star  just  apparently  double  ?  * 

§  52.    Drawing  to  Scale. 

A  straight-line  diagram  to  scale  is  one  similar  to  the 
object  represented,  that  is,  having  all  its  angles  the  corre- 
sponding angles  of  the  object,  and  any  pair  of  its  sides 
bearing  to  each  other  the  same  ratio  as  do  the  correspond- 
ing lines  of  the  object.  Architects'  plans  of  buildings,  sur- 
veyors' plots  of  fields,  are  familiar  examples.  The  ratio 
of  any  line  of  the  drawing  to  the  corresponding  line  of 
the  object  is  the  scale  of  the  diagram.  This  scale  should 
be  indicated  on  the  drawing.  This  may  be  done  by  noting 
on  the  map,  "scale  1  to  10,"  or  "scale  1  inch  to  1  mile," 

*  This  angle  is,  for  most  very  good  eyes,  about  2 J'.  You  can  test  your 
eye  by  seeing  at  which  corner  of  the  parallelogram  under  the  bright  star 
Vega  there  is  a  double  star.  The  stars  are  about  2|'  apart,  and  but  few 
students  can  correctly  report  the  corner.  This  should  convince  the  student 
that  when  he  calculates  seconds,  tenths  of  seconds,  and  even  hundredths  of 
seconds,  in  angles,  as  many  text-books  do,  he  is  splitting  hairs  very  fine. 
Visibility  varies  with  the  eye  and  with  contrast  in  color  of  object  and  back- 
ground, running  from  about  30"  to  2|'.  Mr.  F.  L.  O.  Wadsworth  has  shown 
that  much  of  the  "fine  measurement"  with  instruments  is  merely  an  optical 
delusion. 


§52]  DIAGRAMS   TO   SCALE.  79 

as  the  case  may  be.  Look  this  matter  up  on  some  railway 
map,  or  map  in  a  geography,  and  report  as  to  how  the 
scale  was  indicated.  The  scale  will,  of  course,  depend  on 
the  size  of  the  object  as  compared  with  the  size  of  the  draw- 
ing desired.  It  may  be  an  inch  to  many  miles,  or  an  inch 
to  a  few  feet,  according  to  the  amount  of  detail  desired. 
The  drawing  may  be  on  reduced  scale,  as  railway  maps,  or 
on  enlarged  scale,  as  in  drawings  of  microscopic  objects. 

EXERCISES. 

(Using  protractor  and  straight  edge.) 

1.  Draw  to  scale  a  triangle,  given  two  sides  and  the  included  angle, 
the  sides  being  217,  250,  angle  63°  15',  the  longest  side  of  the  diagram 
not  to  be  over  5  inches.     Measure  the  other  two  angles.     Test. 

2.  Given  two  sides  and  the  angle  opposite  one  of  them.  Sides,  240, 
224;  angle,  47°  30',  opposite  224.  How  many  triangles  are  possible? 
What  change  would  be  necessary  in  one  of  the  given  sides  to  make  just 
one  triangle  possible  ?  Can  the  sides  be  so  given  as  to  make  the  triangle 
impossible  ?  How  many  triangles  are  possible  when  the  angle  given  is 
to  be  opposite  the  larger  of  the  two  sides? 

3.  Given  two  angles  and  the  included  side.  Angles,  30°,  85°  30' ; 
side,  10  feet.    Make  the  diagram  to  scale. 

4.  Given  three  sides,  5.2,  8.2,  9.3.    Make  the  diagram  to  scale. 

5.  Given  the  three  angles.     How  many  triangles  ? 

6.  In  all  the  preceding  cases,  use  the  scale  and  protractor  to  deter- 
mine the  ungiven  parts.  Do  you  know  of  any  way  by  which  you  can 
test  your  results  ? 

7.  Plot  to  scale  a  right-angled  triangle  of  which  one  side  is  20  miles, 
and  one  angle  57°  30'. 

8.  Plot  to  scale  a  right-angled  triangle,  one  of  whose  angles  is  76°  30', 
and  whose  hypothenuse  is  500  feet. 

9.  Plot  to  scale  a  right-angled  triangle  whose  two  legs  are  15  and  12. 

10.  Measure  the  ungiven  parts  in  the  preceding  right-angled  triangles, 
and  from  these  measurements  and  the  scale  calculate  their  values.  Test 
the  same  by  other  calculations. 

11.  Look  up  the  map  of  your  state,  in  some  geography  or  wall  map, 
and  calculate  from  the  given  scale  of  the  map  the  air-line  distances  be- 
tween its  five  largest  cities. 

12.  Calculate  to  three  decimal  places  the  ratio  of  the  sides  of  each  of 
the  right-angled  triangles  in  Exs.  7,  8,  9,  and  the  ratio  of  each  side  to 
the  hypothenuse. 


80  PLANE   TRIGONOMETRY.  [§53 

13.  Plot  a  four-sided  field  whose  sides  are,  in  feet,  500,  240,  120,  180, 
and  whose  angles  are,  following  the  order  of  sides,  80°,  70°,  110°,  A. 
How  large  is  A  ? 

14.  Draw  the  diagonals  in  Ex.  13,  and  calculate  from  the  scale  their 
lengths.  Measure  in  the  diagram  all  angles  not  given,  and  test  them 
by  the  three  angle  tests  which  hold,  together  with  the  corner  tests. 

§  53.    The  Two  Half  Terminals,  Three  Third  Terminals,  etc. 

If  the  angle  A  is  the  principal  angle  of  a  given  terminal, 
this  terminal  is  also  located  by  the  angles, 

A,  A  ±360°,   A  ±720°,   etc. 

Thus  the  half  angles  corresponding  to  a  given  terminal  are 

|,   |  ±180°,    |  ±360°,   etc. 

These  locate  two,  and  only  two,  terminals.     These  terminals 
are  180°  apart. 

Similarly,  one-third  of  the  angle  locating  a  given  terminal 
will  fall  under  the  forms, 

|,    |  ±120°,  |  ±240°,  etc. 

These  locate  three,  and  only  three,  terminals,  120°  apart. 

Similarly,  there  are  four  terminals  for  one-fourth  of  the 
angle  corresponding  to  a  given  terminal,  and  so  on. 

EXERCISES. 

1.  Use  the  protractor  to  draw  the  terminals  for  the  half  angles  cor- 
responding to  the  initial  line. 

2.  Do  the  same  for  the  terminals  corresponding  to  90°,  180°,  270°. 

3.  Show  that  if  one  vertex  of  an  equilateral  triangle  is  on  the  initial 
line,  the  other  vertices  are  on  the  terminals  corresponding  to  one-third  of 
the  angle  of  the  initial  line. 

4.  Show  the  corresponding  proposition  for  the  square,  regular  penta- 
gon, regular  hexagon,  etc. 

5.  If  one  vertex  of  an  equilateral  triangle  is  on  the  terminal  of  180°, 
show  that  the  other  vertices  are  on  the  terminals  of  the  third  of  the 
general  angle  corresponding  to  the  terminal  of  180°. 


§54]  DIAGRAMS   TO   SCALE.  81 

6.   Locate  the  terminals  for  one-quarter  of  the  general  angle  corre- 
sponding to  the  terminal  of  180°;  for  one-fifth;  for  one-sixth. 

Note.  —  Later  it  will  be  shown  that  the  geometric  operations  above 
correspond  to  the  algebraic  solution  of  the  equations, 

x2  =  ±l,   xs  =  ±1,   z4  =  ±l,   x»  =  ±l,  etc., 

by  De  Moivre's  theorem. 


§  54.  Instruments  for  Measuring  Angles. 

It  is  advisable  that  along  with  the  study  of  trigonometry  the  student 
be  given  field-work  in  measuring  angles  with  the  surveyor's  compass, 
with  a  sextant,  and  with  a  transit.  In  no  other  way  can  he  learn  the 
limits  of  accuracy  of  measurement,  and  the  folly  of  calculating  hun- 
dredths of  seconds  when  he  ought  not  to  do  it. 


EXERCISES. 

1.  Investigate  and  report  on  the  following  topics :  (a)  The  accuracy 
of  the  outfit  and  processes  of  a  county  surveyor  in  your  state,  (b)  The 
same  for  a  city  surveyor,  (c)  The  same  for  a  steam  railroad  surveyor. 
(d)  The  same  for  the  surveyor  of  an  electric  railway,  (c)  The  means 
of  measuring  small  angles  in  engineering  and  astronomy  and  the  small- 
est readings  in  different  sorts  of  surveying.  (/)  Measuring  base-lines  in 
geodetic  surveys,     (g)  Limits  of  accuracy  in  government  land-surveys. 

2.  A  city  lot,  about  29  feet  by  290  feet,  was  sold  at  $1000  a  square 
foot,  on  the  measurements  29.3  feet  by  293.7  feet.  Show  that,  possibly, 
the  owner  lost  about  $14,600,  by  not  having  the  short  side  measured 
with  the  same  relative  accuracy  as  the  long  side.  Which  would  entail 
the  larger  loss,  dropping  .3  on  the  short  side  or  .7  on  the  long  side? 

3.  Show  that  J^^  =  log6  a.     Show  that  ax  =  exl0^°,  and  from  the 

loge  b 

series  for  ex  (§8)  deduce  a  series  for  ax. 

4.  Show  that  if  a  set  of  numbers  form  a  G.  P.,  their  logarithms  form 
an  A. P.  Show  that  the  arithmetic  mean  of  two  logarithms  is  the  loga- 
rithm of  the  geometric  mean  of  the  two  numbers  corresponding  to  the 
given  logarithms.  Show  that  the  logarithm  of  any  number  can  be  cal- 
culated to  any  desired  degree  of  accuracy  by  the  continued  insertion  of 
arithmetic  and  geometric  means.  Calculate  logio2  to  four  figures  in 
this  way. 


CHAPTER  V. 

THE    SINE,    ANTI-SINE,    RECIPROCAL    SINE    (COSECANT), 
AND    COVERSED    SINE   OF   AN   ANGLE. 


§  55.   Coordinates. 

(a)  Rectangular  Coordinates  and  Rectangular  Coordinate 
Paper.  Draw  through  0  a  pair  of  mutually  perpendicular 
lines,  giving  the  quadrants  as  explained  in  §  45.     0  is  called 

the  origin  of  coordinates,  also  the 
origin,  also  the  pole.  The  border 
lines  of  the  quadrants  are  called 
the  axes  of  coordinates.  And  for 
distinction,  the  old  initial  line 
(right  and  left)  is  called  the 
axis  of  abscissas,  and  the  verti- 
cal line  the  axis  of  ordinates. 
From  any  point,  P,  in  the  termi- 
nal OP,  of  a  given  angle  A  OP, 
drop  a  perpendicular  PM,  on 
the  initial  line  OA  (produced  if 
necessary).  Then  OM  is  called  the  abscissa  of  the  point  P, 
and  is  indicated  by  the  letter  x ;  MP,  its  ordinate  or  y ;  and 
OP  its  modulus,  and  also  its  radius  vector  (we  shall  use  the  word 
modulus  and  the  letter  r).  Notice  particularly  the  order  in 
which  these  lengths  are  written :  It  is  OM,  not  MO ;  it  is  MP, 
not  PM;  it  is  OP,  not  P  0.  This  is  of  primary  importance, 
for  reversal  of  the  order  in  which  a  length  is  written,  when 
sign  is  considered,  is  a  reversal  of  direction  and  equivalent 
to  a  change  of  sign.  As  to  sign,  it  is  now  universally  agreed 
that  all  lines  taken  upward,  no  matter  where  starting,  are  +  ; 

82 


T 

S 

^  p 

4-                                   J,- 

^          X-         5* 

s                   r? 

ss^          *f 

^       r»             J? 

X                SO     S    ±         ]lf. 

M                   \/                       M 

M            ^'v^               M 

^                    ^^ 

7                           ^^ 

S                               ^-L 

^                                   "    £_ 

^P 

f 

Fig.  13. 


§  55]  THE   SINE  FAMILY.  83 

downward,  — ;  to  the  right,  -f ;  to  the  left,  — .  Where  no 
sign  is  stated,  +  is  understood.  A  point  in  the  plane  of  the 
axes  is  located  by  giving  its  coordinates  in  a  parenthesis,  as 
(2,  3),  the  abscissa  being  in  the  lead.  The  point  (2,  3)  is 
two  scale  units  to  the  right  of  the  vertical  axis  and  three 
scale  units  above  the  horizontal  axis.  To  reach  it,  go  two 
units  from  0  along  the  axis  of  abscissas  to  the  right;  then 
three  units  vertically  up  from  the  point  so  reached  on  the 
abscissa  axis.  Similarly,  to  locate  (—2,  3),  go  two  units 
along  the  abscissa  axis  to  the  left  from  0,  and  then  three 
units  vertically  up  from  the  point  so  reached.  How  are 
(2,  -  3),  and  (-2,-3)  located  ? 

The  teacher  may  select  at  random  sufficient  practice  examples  to  make 
sure  that  the  matter  is  understood  by  the  class. 

In  quadrants  I  and  II,  ordinates  are  -+- ;  in  III,  IV,  — . 
In  quadrants  IV,  I,  abscissas  are  -f- ;  in  II,  III,  — . 

The  modulus  OP,  of  the  point  P,  is  always  counted  plus 
when  it  lies  along  the  terminal  of  the  given  angle.  For  all 
points  in  the  opposite  terminal  (the  terminal  continued  back- 
wards through  the  origin),  it  is  minus.  Thus,  while  OP 
might  be  counted  plus  for  a  certain  angle,  this  same  line 
OP  would  be  minus  for  the  angle  which  is  180°  more  than 
the  given  angle,  or  any  odd  integral  multiple  of  180°  more 
than  the  given  angle.  Even  when  the  terminal  falls  on 
the  left-hand  horizontal  line  or  on  the  downward  vertical, 
the  modulus  for  any  point  on  it  is  counted  plus,  while  in  the 
first  case  the  ordinate  is  zero  and  the  abscissa  negative ;  and 
in  the  second,  the  reverse.  Any  measurement  along  the  termi- 
nal, away  from  the  origin,  is,  by  agreement,  plus  always,  no 
matter  whether  the  angle  is  plus  or  minus,  and  no  matter  what 
its  size. 

Paper  like  that  on  which  Fig.  13  is  constructed  is  called  rectangular 
coordinate  paper.  It  can  be  bought  in  small  sheets  or  by  the  yard,  and  of 
various  divisions  to  the  inch  (or  centimeter).  That  10  x  10  to  the  inch 
is  convenient  for  locating  points  in  a  drawing. 


84 


PLANE   TRIGONOMETRY. 


[§  55 


(b)  Polar  Coordinates  and  Polar  Coordinate  Paper.  A 
point  on  a  plane  is  also  located  by  giving  the  length  of  its 
modulus  and  the  angle  which  the  modulus  makes  with  the 
initial  line.  These  are  given  in  a  parenthesis  as  (r,  0),  the 
modulus  being  set  first.  By  the  previous  agreement  as  to 
sign  of  the  modulus,  a  point  indicated  by  (2,  30°)  would 
be  two  units  from  the  origin  along  the  terminal  of  30°, 
while  ( —  2,  30°)  would  be  two  units  along  the  opposite 
terminal  from  the  origin.  Where  will  (2,  —  30°)  and 
(_2,   -30°)  fall? 

EXERCISES. 

1.   Calling  the  small  divisions  on  radii  units,  locate  on  Fig.  14  the 

points  (4,  5°),  (5, 
5°),  (6, 5°),  (4,  2(F), 
(4,  25°),  (4,  50°), 
(4,90°),  (-5,5°), 
(-5,  50°),  (-5, 
180°),  (-5,275°), 
(-8,  -20°),  (-8, 
-80°),  (-8, -180°), 
(-8,  -270°),  (-8, 
3000°). 

2.  Similarly  lo- 
cate on  Fig.  15  the 
following  points, 
the  second  figure 
in  each  parenthesis 
being  radian  meas- 
ure: (6,3),  (7,3), 
(8,  3),  (8,  4.3),  (9, 
5.5),  (6,  6.3),  (7, 
-  0.5),  (4,  -  0.6), 
(4,3.14159),  (4,5, 
4, 


Fig.  14.* 


0, 


■),  (-4,  -*■),  (-4,  -|),  (-4,  -|),  (-4, 


3),  ( 


6). 


3.    Construct  to  scale  the  following  points  in  rectangular  coordinates, 
using,  preferably,  rectangular  coordinate  paper  :  (2,  3),  (2,  —  3),  (  —  2,  3), 


*  Figure  14  is  in  5°  angle-spaces ;  Fig.  15  is  in  tenths  of  radians.     The 
divisions  on  the  radius  are  in  each  case  arbitrary. 


§55] 


THE   SINE   FAMILY. 


85 


(-  2,  -  3),  (7,  8),  (15,  -  3),  (-  5,  4),  (-  9,  -  3),  (3,  -  3),  (3,  -  1),  (150, 
160),  (900,  -800),  (3.3,  4.1),  (-3.3,  -  4.1),  (5.6,  -  7.3). 

4.  Where  are  (0,  0),  (0, 1),  (-1,  0),  (1,  0),  (1,  0),  (0,  -  1)? 

5.  Use  scale  and  protractor,  or  Groat's  coordinate  paper,  to  construct 
the  following  points:    (2,  30°),    (2,   -30°),    (-2,  30°),    (-2,  -30°), 

(2,  t),  (3,  60°),  (3,  -  60°),  (-  3,  -  45°),  (3,  45°),  (5,  -  tt),  (-  5,  -^*j, 
(5,  =j£),  (150,2*-),  (l50,  ^),  (2,  lr),  (-  2,3-),  (-  2,  -  1'),  (2,  -3'). 

6.  In  what  quadrant  is  a  point  whose  abscissa  and  ordinate  are  both 
plus?    Both  minus? 

Abscissa  plus,  or- 
dinate minus?  Or- 
dinate plus,  abscissa 
minus  ? 

7.  Use  Groat's 
coordinate  paper, 
taking  the  small 
divisions  on  the  radii 
as  units,  to  locate  to 
the  extent  of  the 
sheet  the  points  (ra- 
dian measure  for  an- 
gles):  (0,0),  (1,0.1), 
(2,  0.2),  (3,  0.3), 
(4,  0.4),  etc.  Then 
join  all  these  points 
by  a  smooth  curve. 
This  will  represent 
a  particular  form  of 
the  Spiral  of  Archimedes,  namely, 

r  =  lO0, 

the  general  curve  being  r  =  aO.     Note  that  in  each  point  above  r  is  10 
times  6,  unit  for  unit. 

8.  Use  rectangular  coordinate  paper,  taking  the  small  divisions  as 
units,  to  locate  the  points :  (0,  0),  (1,  1),  (2,  2),  (3,  3),  (4,  4),  (5,  5),  etc., 
(-  1,  -1),  (-  2,  -  2),  (-  3,  -  3),  (-  4,  -  4),  etc.  Join  these  points. 
What  then  is  y  =  x  ? 

9.  Use  rectangular  coordinate  paper  to  locate  (0,  2),  (1,  5),  (2,  8), 
(3,  11),  (4,  14),  where  y  is  always  3x  +  2.     Join  these  points.    What  do 


Fig.  15. 


86 


PLANE   TRIGONOMETRY. 


[§56 


you  get?  What  are  some  negative  points  on  the  same  figure?  What, 
then,  does  y  =  3  x  +  2  represent?  What  does  y  =  mx  +  b  represent,  if  m 
and  b  are  fixed  numbers  ? 

10.   If  a  point  is  located  by  (r,  A0),  what  variety  of  form  may  r  and 
A0  take  to  locate  one  and  the  same  point? 


§  56.    The  Sine  of  an  Angle. 


The  sine  of  an  angle 


1 

> 

/ 

/ 

/ 

/ 

/ 

t 

/ 

) 

/ 

& 

/ 

I 

1 

I 

- 

the  ratio  of  the  ordinate  of  any  point 
on  the  terminal  of  the  angle  to 
the  modulus  of  the  same  point,  or, 
in  Fig.  16, 

MP 


Sine  of  6  = 


OP 


Fig.  16. 


This  is  known  as  the  ratio- 
definition  of  the  sine,  or  as 
Hassler's  definition,  to  distin- 
guish it  from  the  so-called  line- 
definition.     See  §  67. 

It  must  be  considered  more 
as  a  working  definition  than 
as  one  of  theoretic  exactness.  Sines  are  not  calculated  by 
laying  out  the  angle  with  a  protractor,  measuring  ordinate 
and  modulus  and  then  dividing  the  former  by  the  latter. 
Later  it  will  be  found  that  connected  with  every  angle  6, 
expressed  in  circular  measure,  is  a  certain  number,  the  sine 
of  the  angle,  which  can  be  calculated  for  any  given  value  of 
6  by  the  following  series,  to  any  desired  degree  of  exactness : 

Sin  0=6  —  —  +—  —  —  ...,  etc.,  to  infinity. 

This  series  gives  the  correct  theoretic  definition  of  the  sine  of 
the  angle  6.  The  sine  of  an  angle  is  thus  a  number  connected 
with  the  angle  in  a  specific  manner,  as  indicated  by  the  above 
series,  and  calculable  when  the  angle  is  given. 

These  numbers  have  been  calculated,  approximately,  and 
tabulated  for  certain  angles,  forming  what  is  called  a  Table 
of  Natural  Sines. 


§56]  THE  SINE   FAMILY.  87 

All  that  Hassler's  definition  means  is  that  if  the  ordinate 
were  measured  with  exactness  and  the  modulus  with  exactness, 
and  then  the  division  of  the  corresponding  numbers  carried 
out  as  indicated,  there  would  be  obtained  a  number  agreeing 
with  that  obtained  from  the  series  for  the  same  angle. 

Calculations  of  the  sine  from  the  series  give  the  sine  in 
the  form  of  a  decimal  fraction.  Since  the  series  has  an  infi- 
nite number  of  terms,  the  sine  of  an  angle  can  be  calculated, 
as  a  rule,  only  approximately.  The  tables  give  the  sines  to 
3,  4,  5,  6,  7, 10  places.  A  table  of  sines  giving  them  to  three 
places  of  decimals  is  called  a  three-place  table;  one  giving 
them  to  four  places  of  decimals,  a  four-place  table,  and  so  on. 

Ordinates  and  moduli  and  angles  can  be  measured  only 
approximately.  Consequently,  if  the  sine  of  a  measured 
angle  were  obtained  from  Hassler's  definition  and  the  result 
compared  with  the  tables,  the  closeness  of  agreement  would 
depend  upon  the  degree  of  accuracy  with  which  the  measure- 
ments of  lines  and  angle  had  been  carried  out. 

The  practical  value  of  Hassler's  definition  is,  therefore, 
this :  It  indicates  that  if  the  ordinate  and  modulus  of  an 
angle  in  a  diagram,  or  in  the  field,  are  measured,  and  their 
ratio  obtained  and  compared  with  the  tables,  the  tables  will 
show  an  angle  (one  of  many)  which  is,  approximately,  the 
angle  of  the  diagram,  or  one  from  which  that  angle  can  be 
determined,  approximately.  The  closeness  of  the  approxi- 
mation will  depend  upon  the  accuracy  with  which  the  meas- 
urements are  made. 

The  sine  of  an  angle  is  thus  a  number  connected  with  the 
angle  and  having  under  Hassler's  definition  a  threefold  prac- 
tical value : 

(1)  When  ordinate  and  modulus  are  given,  it  indicates  the 
angle. 

(2)  When  the  angle  and  modulus  are  given,  it  is  the  number 
which  multiplied  by  the  modulus  will  give  the  ordinate. 

(3)  When  the  angle  and  ordinate  are  given,  it  is  the  number 
by  which  the  ordinate  is  divided  to  give  the  modulus. 


88  PLANE   TRIGONOMETRY.  [§56 

These  three  uses  to  which  the  sine  may  be  put  constitute 
its  claim  to  importance  as  a  calculation  device  and  give  to 
Hassler's  definition  its  value. 

LABORATORY  EXERCISES. 

That  the  student  may  get  firmly  fixed  in  his  mind  the  definition  of 
the  sine  as  a  ratio,  the  teacher  may  here  show  him  how  to  look  up  in  the 
tables  the  sines  of  angles  in  degrees  (omitting  minutes  and  seconds). 
The  student  may  then  lay  out  to  scale  some  five  such  selected  angles, 
measure  the  corresponding  ordinate  and  modulus,  divide  (to  one  or  to 
two  figures),  and  compare  with  the  tables. 

He  may  also  take  the  modulus  one  inch  long  and  show  that  the 
ordinate  (in  decimals  of  an  inch)  is  the  table-sine.  Do  the  same  with  a 
modulus  ten  inches  long.    Make  ten  such  constructions  and  measure. 

EXERCISES. 

(Make  diagrams  to  scale.) 

1.  If  tiie  modulus  is  5  and  the  ordinate  3,  what  is  the  sine  of  the 
corresponding  angle  ?  Calculate  the  sine  to  two  decimal  places  and  get 
the  angle  when  the  modulus  is  34  and  the  ordinate  25 ;  to  three  places 
with  ordinate  25.3  and  modulus  34.7  ;  to  four  places  with  ordinate  25.37 
and  modulus  34.72. 

2.  The  side  of  a  square  is  V2  inches ;  calculate  to  four  decimal  places 
the  sine  of  the  angle  which  the  diagonal  makes  with  the  side,  and  com- 
pare with  the  tables.  Show  that  the  sine  of  the  angle  which  the  diagonal 
makes  with  the  side  is  independent  of  the  length  of  the  side. 

3.  Assuming  the  side  of  an  equilateral  triangle  as  2,  calculate  the  sine 
of  60°  to  four  decimal  places.     Do  the  same  when  the  side  is  a. 

4.  By  drawing  an  equilateral  triangle  whose  side  is  2,  so  that  its 
median  is  horizontal  and  the  initial  line,  calculate  to  four  decimal  places 
the  sine  of  30°.     Compare  the  result  with  the  table-value. 

5.  Give  the  sines  of  30°,  45°,  60°,  in  the  form  of  radicals. 

6.  The  modulus  for  an  angle  whose  sine  is  0.35  is  27,  what  is  the 
ordinate  ?  How  high  up  on  the  side  of  a  house  will  a  ladder  50  feet  long 
reach  when  tilted  at  an  angle  of  30°  with  the  ground?  45°?  60°?  How 
high  when  the  sine  of  the  angle  of  tilt  is  0.42  ?  Give  all  these  results  to 
only  two  figures. 

7.  The  ordinate  for  an  angle  whose  sine  is  0.24  is  53,  what  is  the 
modulus  ?    If  the  roof  of  a  house  is  inclined  to  the  horizontal  at  an  angle 


58] 


THE   SINE   FAMILY. 


89 


of  30°  and  the  ridge-pole  is  15  feet,  how  long  are  the  rafters?  How  long 
when  the  angle  is  45°  ?  60°  ?  How  long  when  the  sine  of  the  angle  is 
0.32  ?    Give  results  to  only  two  figures. 

8.  The  student  will  devise  seven  other  examples  like  the  above,  solve 
them  and  hand  them  in.  Let  five  of  them  be  practical  examples  (roofs, 
etc.)  within  the  student's  experience. 

§  57-  The  sine  has  many  interesting  and  important  properties,  aside 
from  its  use  in  calculations.  All  these  can  be  deduced  directly  from  the 
series  definition,  without  reference  to  Hassler's  definition.  In  fact,  a 
text-book  could  be  based  on  the  series  definition.  However,  for  that  the 
times  seem  not  yet  ripe. 

§  58.  Hassler's  definition  holds  for  all  positions  of  the 
terminal,  no  matter  in  which  quadrant  it  falls. 


T                X 

X               ST 

T                x 

Z>U                                         \ 

%S                s^ 

S                   s 

y                      s 

*                                                                                          Sr- 

_,•                                 s 

Q_<S_-         _  _                          _SV0 

1                                                      Jtf.                                 J\f 

±  ■                                                 ± 

Fig.  17. 


Fia.  18. 


M 


/ 


<P 


/ 


M 


m 


Fig.  19. 


Fig.  20. 


90  PLANE   TRIGONOMETRY.  [§58 

In  each  of  the  preceding  diagrams  (Figs.  17-20),  corre- 
sponding to  the  four  quadrants,  if  0  is  any  one  of  the  angles 
corresponding  to  any  one  of  the  indicated  terminals, 

0.         p  a      ordinate  MP      y 

Sine  of  6  = — — -=  *« 

modulus  OP       r 

It  is  customary  to  contract  sine  of  6  to  sin  6.  It  is  read, 
"sine  0,"  omitting  "of." 

§  59.    That  the  value  of  the  sine  is  independent  of  the 
length   of    the   modulus,   is   apparent    at    once    from    the 
accompanying  diagram  (Fig.  21). 
pl 
F^\        Tne  triangles  OMP,  01M1P1  are  similar. 

MP     MjPt 

OP       OPx' 

The  sine  of  an  angle  is  thus  a  ratio,  or  number,  whose  value 
depends  only  on  the  position  of  the  terminal  line  as  related 
to  the  initial  line.  It  refers  not  so  much  to  a  specific  angle 
as  to  a  specific  position  of  the  terminal. 

LABORATORY    EXERCISE, 

Lay  out  an  angle  of  18°  with  the  protractor.  Select  five  different 
moduli  (one  of  them  an  inch).  Measure  the  moduli  and  ordinates. 
Divide,  compare  results  with  each  other  and  with  the  table.  Carry  out 
the  divisions  only  so  far  as  the  means  of  measurement  used  justify. 

All  ratios  are  numbers.  They  are  sometimes  called  pure 
numbers  to  distinguish  them  from  the  so-called  denominate 
numbers.  Some  special  remarks  on  ratios  may  be  appropriate 
here. 

§  60.    General  Eemarks  on  Ratios.    The  Sine  a  Continuum. 

Ratios  among  real  numbers  are  of  three  kinds : 
(a)  Positive  or  negative  integers. 
(£>)  Positive  or  negative  common  fractions, 
(c)   Unending,  non-repeating  decimals,  positive  or  negative. 


§60]  THE   SINE   FAMILY.  91 

Ratios  of  the  second  kind,  (6),  are  of  two  classes : 

(i)    Ending  decimals,  like  J  =  0.5,  ^  =  0.2,  etc. 

(ii)  Repeating  decimals,  like  i  =  0.333  •••,  J  =  0.1666  •••. 

All  common  fractions  whose  denominators  are  2  or  5,  or 
powers  of  2,  or  of  5,  or  of  2  and  5,  will  form  ending  decimals. 
For  example : 

1  =  0.5;  £  =  0.2;  J  =  0.25;  £  =  0.04,  etc. 

All  other  common  fractions,  when  expressed  decimally, 
give  repeating  decimals.  This  will  become  evident  by  con- 
sidering some  special  example,  as  ^f  1,  and  carrying  out  the 
division  as  follows,  setting  the  remainders  in  the  upper  hori- 
zontal line : 

46.451326 

7)251.000000 

35.857142 

The  remainder,  when  the  zeros  first  come  into  use,  is  6. 
Since,  now,  in  dividing  by  7,  there  can  be  only  six  different 
remainders,  and  since  the  division  is  to  be  non-terminating, 
this  special  remainder,  6,  appearing  in  connection  with  the 
use  of  the  added  zeros,  must,  sooner  or  later,  reappear,  as 
shown  in  the  actual  division.  As  soon  as  this  remainder 
reappears,  there  will  be  a  repetition  of  the  quotient  figures 
obtained  since  the  6  first  appeared  as  remainder.  What  has 
happened  here  with  7  is  readily  seen  to  be  general.  For 
with  any  other  denominator,  as,  say,  213,  there  can  be  only 
212  different  remainders  for  non-terminating  division,  and 
that  one  present  when  the  zeros  are  first  used  must  again 
reappear,  giving  a  repetition  of  quotient  figures. 

EXERCISES. 

Show  that  the  following  fractions  form  repeating  decimals : 


92  PLANE   TRIGONOMETRY.  [§60 

Examples  of  numbers  of  the  third  class,  (V),  the  unending 
non-repeating  decimals,  are : 

7r  =  3.14159  •••,  the  ratio  of  the  circumference  of  a  circle 
to  its  diameter. 

e  =  2.7182818  •••,  the  base  of  the  Napierian  logarithms. 

V2  =  1.41~., 

and  all  other  real  numbers  which  do  not  belong  to  classes 
(a)  and  (5).  Originally,  only  the  roots  (of  rational  num- 
bers) which  could  not  be  extracted  exactly  were  called 
the  irrational  numbers.  Since,  however,  there  is  no  essen- 
tial difference  between  such  roots  and  numbers  like  7r,  e, 
in  so  far  as  being  neither  ending  nor  repeating  is  con- 
cerned, all  real  numbers  not  of  classes  (V),  (5),  may  be 
called  irrational. 

If  the  numbers  of  (a),  the  positive  and  negative  integers, 
are  plotted  on  a  straight  line,  as  in  algebra,  they  are  repre- 
sented by  separate  points,  a  unit  distance  apart.  The  num- 
bers of  (6)  when  plotted  on  the  same  straight  line  help  to  fill 
in  the  points  between  those  occupied  by  the  numbers  of  (a) ; 
but  (a)  and  (5)  do  not  occupy  completely  all  the  points  on 
the  straight  line.  Class  (<?)  comes  in  to  take  up  the  remain- 
ing points.  The  complete  set  of  numbers  of  classes  (V), 
(6),  (<?),  occupy  the  line  completely,  so  that  each  number 
occupies  a  point,  and  each  point  represents  a  number,  pas- 
sage from  point  to  point  along  the  line  representing  the 
growth  of  one  number  into  another.  Such  a  set  of  numbers 
is  called  a  continuum,  in  contradistinction  to  the  discreet  set 
of  numbers,  the  units,  or  the  units  and  common  fractions,  or 
any  set,  or  pairs  of  sets,  of  the  numbers  (a),  (3),  (<?). 

The  sines  of  angles  take  up,  as  will  be  seen  later,  but  a 
very  limited  portion  of  this  continuum,  namely,  that  from 
+  1  to  —  1,  these  limits  included ;  but  they  occupy  this 
limited  region  completely,  forming  themselves  a  continuum 
(§  69,  «). 


§62] 


THE   SINE    FAMILY. 


§  61.    The  Sign  of  the  Sine. 

Since  the  modulus  is  always  plus  along  the  terminal  of  an 
angle  (§  '55),  the  sign  of  the  sine  is  that  of  the  ordinate. 

.'.  All  angles  of  whatever  size  or  sign, 
with  terminals  in  quadrants  I,  II, 

have  positive  sines ; 
with  terminals  in  quadrants  III, 

IV,  negative  sines.  FIQ.  22. 


Quadrant 

I 

II 

III 

IV 

Sine 

+  - 



EXERCISES. 

(Make  diagrams  to  scale.) 

1.  What  signs  have  the  sines  of  the  following  angles  (degrees): 

30,  -  30,  135,  -  135,  185,  -  185,  275,  -  275,  361,   -  361,  455,  -  455 
570,  -  570,  650,  -  650,  1700,  -  1700. 

2.  Determine  the  signs  of  the  sines  of  the  following  angles ; 

7T      7T   2?T      2lT       4:7T  4  7T   7  7T      7  7T   7  7T      7  7T   7T      7T 


3'        3'     3  '         3  '     3  '     ~  3  '     4  '     ~  4  '     3  ' 
7tt     _7tt     17tt      _17tt 
5  '    "  ~5~'    "d"'        "6"' 


3  '    6'        6' 


§  62.   Angles  with  the  Same  Sine. 
The  sine  for  the  terminal  position  OP  (Fig.  23)  is 


MP 
OP' 


•\  (i)  All  angles  with  the  same  terminal  have  the  same 
sine. 


s 

^ 

^.y** 

-,^        ^ 

4^ 

Jr 

%    -p                                  t    *y 

s^«                                 n^ 

%                             S 

V                                             >* 

5fc                             S 

32  S^     +<L& 

-At     ^^T^ 

M                         O                    M 

Fia.  23. 


Fio.  24. 


94 


PLANE   TRIGONOMETRY. 


[§62 


Taking  6  (circular  measure),  A0  (Fig.  24)  as  the  principal 
angle  of  OPx  (see  §  46),  the  angles  of  this  terminal  are 

2nir  +  6,   or   2 ?i  •  180°  +  .A°. 
.-.  sin  (2  mr  +  0)  =  sin  #, 
or,  sin  (2  n  •  180°  +  A°)  =  sin  A°, 

where  n  is  any  positive  or  negative  integer. 

(ii)  Terminals  symmetrically  inclined  to  the  vertical,  as 
OPv   OP2  (Fig.  24),  or  as  OP3,  OPi  (Fig.  25),  belong  to 

angles  with  the  same  sine,  since 
for  such  angles  equal  terminals 
give  equal  ordinates. 

The  angles  of  the  terminal 
OM1  (Fig.  24)  are  2mr  or 
2n- 180°,  and  those  of  OM2  are 
(2w  +  l>,  or  (8*4-1)  -180°. 
The  position  OPx  (Fig.  24)  is 
best  reached  by  going  forward 
the  principal  angle  #,  or  A°, 
from  the  terminal  OM1 ;  the 
position  OP2,  by  going  back- 
ward the  same  angle  from  the  position  OM2. 

.\  sin  {  (2  n  +  1)tt  -  (9}  =  sin  (9, 
or,  sin  {(2  n  +  1)180°  -  .A°}  =  sin  4°. 

In  particular,  sin  (jr  —  0)  =  sin  0, 

and  sin  (180°  -  A0)  =  sin  A°. 

(iii)  Any  angle  of  terminal  OP2  (Fig.  24)  has  the  same 
sine  as  any  angle  of  terminal  OPv 

.'.  sin{(2w  +  l>r-0}  =  sin(2wwr  +  0), 
or,        sin  j  (2  n  +  1)180°  -  A° \  =  sin  (2  m- 180°  +  4°), 

where  m,  w  are  any  positive  or  negative  integers. 

(iv)  Thus,  in  general,  all  angles  having  the  same  sine  as 
0  or  A°,  fall  under  the  forms 


nir 

M'i 

C                    AU 

S,    •  -</> 

'  i~4%7 

S^T 

*_ 

x 

y 

s 

S 

*». 

S 

s^ 

7 

s 

7 

s 

o 

-J* 

srS 

2mr  +  9 

(2n  +  l)ir 


I   0r   l(2w+l)180°-^°> 


§62]  THE  SINE  FAMILY.  95 

Since  (—  V)m  is  +  1  when  m  is  even  and  —  1  when  m  is  odd, 

the  two  preceding  sets  of   formulas  are  both  included  in 

these  : 

mir  +  (  -  l)«*e, 

m  .  180°  +  (  -  1)™  .  A°, 

where  m  is  any  positive  or  negative  integer. 

(v)  The  general  result  of  (iv)  is  so  important  that  a 
verbal  statement  may  also  be  given: 

If  any  given  angle  in  degree  measure  is  added  to  an  even 
multiple  of  180°,  or  subtracted  from  an  odd  multiple  of  180°, 
an  angle  is  obtained  having  the  same  sine  as  the  given  angle. 

The  student  may  give  the  corresponding  statement  for 
angles  in  radian  measure. 

(vi)  While  in  the  preceding  formulas  any  angle  of  the 

given  terminal  may  be  taken,  it  is  customary  to  take  the 

principal  angle  of  the  terminal.     For  example,  while  187° 

and  — 173°  locate  the  same  terminal,  one  would  use  as  the 

general  expression  for  all  angles  whose  sine  is  that  of  this 

terminal, 

m.180-(-rpl73°, 

rather  than  m  .  180  +  (  -  1)™  187°, 

though  they  both  give  the  same  set  of  values,  when  m  is 
taken  from  —  oo  to  -f-  oo. 

EXERCISES. 

1.  Find  twelve  angles,  six  being  plus  and  six  minus,  having  the  same 
sine  as  30°.    In  which  quadrants  will  these  angles  lie  ?    Give  the  diagram. 

2.  Do  the  same  for  -  60°,  and  for  ?  and  ^JL 

4  4 

3.  Select  sufficient  examples  of  the  same  character  to  fix  the  formula 
in  mind.     Stop  when  you  know  it  thoroughly. 

4.  From  Fig.  25,  show  sin  (imr  +  (-  l)m<£)  =  sin  <f>. 

5.  Show  that  sin (ax  •  bx)  =  sin  (rrnr  +  (-  l)™exl0?eah) 

and  that  sin  —  =  sin  (rrnr  +(—  l)mex  °Sei). 

6*  v  y 


96  PLANE   TRIGONOMETRY.  [§63 

6.  Find  the  angles  which  satisfy  the  equation  sin  9  6  =  sin  8  6. 

All  angles  having  the  same  sine  as  8  6  are  of  the  form  2mr  +  8  0,  or 
of  the  form  (2  n  +  \)tt  —  8  0.  To  satisfy  the  given  equation,  9  0  must 
fall  under  the  one  or  the  other  of  these  forms  : 

.-.   90=2n7r  +  80,  (1) 

or,  90  =  (2n  +  l)7r-80.  (2) 

By  (3),  0  =  2n7r,  and  by  (4),  $  =  (2»  +  1)^. 

The  student  may  make  an  illustrative  diagram  and  note  the  positions 
of  the  terminals. 

7.  Solve  similarly  and  give  diagrams  for 

(a)   sin  7  6  =  sin  0, 
(6)    sin20  =  sin30, 
(c)    sin  w?0  =  sin  r$. 

8.  Make  up  three  examples  like  those  of  Ex.  7,  and  solve ;  illustrate 
by  diagrams. 

§  63.   Angles  with  Opposite  Sines. 

Numbers  which  are  equal  numerically  but  opposite  in  sign 
are  said  to  be  opposite,  or  oppositely  equal. 

(i)  Angles  with  opposite  terminals  have  opposite  sines. 
Here  for  equal  moduli  the  ordinates  are  oppositely  equal. 

.-.  sin  (  ±  180°  +  A°)  =  -  sin  A°, 
sin  (  ±  it  +  0)  aa  —  sin  6. 
And,  in  general, 

sin{(2  n  +  1)  180°  +  A0}  =  -  smA°, 
sin  {(2  n  +  1)  it  +  0}  =  -  sin  9, 

where  n  is  any  positive  or  negative  integer.         ♦ 

(ii)  Terminals  symmetric  to  the  horizontal  give  for  equal 

moduli  oppositely  equal  ordinates.     Thus  all  angles  whose 

terminals   are  symmetric   to  the   horizontal   have    opposite 

sines.     Angles  numerically  equal  but  opposite  in  sign  form 

a  special  case. 

/.  sin  (-0)  =  -  sin0, 

sin  t-A°)  =  -  sin  A°. 


63] 


THE   SINE  FAMILY. 


97 


(iii)  Any  angle  of  a  terminal  has  the  opposite  sine  of  any 
angle  of  the  opposite  terminal. 

sin  \ (2  n  +  1)  ir  +  0\  =  -  sin  (2  mir  +  0), 

or,  sin  j  (2  n  +  1)  180°  +  A°\  =  -  sin  A% 

where  m,  w  are  any  positive  or  negative  integers. 


-£*- 

"7 

y 

s 

*t-                y 

*  -i                 y 

y  o           i  l 

-+  -fe           «^ 

*r 

i? 

jzL 

^f£ 

Fig.  26. 


^ 


\ 


.1/ 


iT\ 


Fig.  28. 


Fig.  29. 


X 

A 

^■^^ 

^^\  0                      ''j 

-M                           ^v». 

^2                                           ^"-^ 

^^\^ 

""•  ^ 

?- 

Fig.  27. 

2j 

"* 

_5k 

^ 

-S*     *r- 

if     -                            ^-              — 

~2 

::e.               j 

..    -*:                      ■  4- 

^                          ■ ' 

"P                                                                                      -T 

*t 

(iv)  Any  angle  of  any  terminal  has  the  opposite  sine  of 
any  angle  of  the  terminal  symmetric  to  the  horizontal. 

sin  (2  nir  +  0)  =  —  sin  (2  mir  —  0), 

sin  (2  n  •  180  +  A°)  =  -  sin  (2  m  180°  -  A0), 

where  m,  w.  are  any  positive  or  negative  integers. 


98  PLANE   TRIGONOMETRY.  [§63 

(v)  Thus,  while  adding  an  angle  in  degree  measure  to  an 
even  multiple  of  180°  or  subtracting  it  from  any  odd  mul- 
tiple of  180°  leaves  the  sine  unchanged,  the  reverse  changes 
the  sign  of  the  sine. 

(vi)  The  student  is  advised  against  attempting  to  memo- 
rize the  preceding  formulas  as  mere  facts.  It  is  better  to 
hold  in  mind  the  effect  on  the  terminal  of  any  given  angle, 
of  its  addition  to,  or  subtraction  from,  given  multiples  of 
180°,  or  of  7r  radians,  noting  that: 

(«)    For  terminals  symmetric  to  the  vertical,  sines  are  equal. 
(/3)  For  terminals  symmetric  to  the   horizontal,  sines  are 
opposite. 

(7)   For  terminals  which  are  opposite,  the  sines  are  opposite. 

EXERCISES. 

1.  Determine  from  a  diagram  five  positive  angles  having  the  opposite 
sine  of  30°,  also  five  negative  angles.  Let  one-half  the  number  of  angles 
be  on  the  opposite  terminal,  the  remainder  on  the  terminal  symmetric  to 
the  horizonal.  Compare  the  results  obtained  from  the  diagram  with  the 
formulas. 

2.  Do  the  same  for  45°;  60°;  120°. 

3.  Do  the  same  for  -135°;   -210°;   -300°. 

4.  Do  the  same  for  1780°;  2117°;  3185°. 

5.  Do  the  same  for  |;  -;  -  ?;  -  |tt. 

6.  What  effect  on  the  terminal  of  an  angle  has  the  addition  or  sub- 
traction of  a  double-even  multiple  of  90°  ?  What  a  double-odd  multiple 
of  90°  ?    Find,  among  the  first  fifty  numbers,  those  multiples  of  -  radians 

which  may  be  added  to  or  subtracted  from  a  given  angle  without  chang- 
ing its  sine.  Also  among  the  second  fifty  numbers,  those  multiples  of 
90°  which  change  the  sign  of  the  sine,  when  added  or  subtracted. 

§  64.    The  Angles  in  a  Table  of  Sines. 

From  the  preceding  sections,  it  is  seen  that  the  sine  of  any 
angle  can  be  connected,  in  equality  or  opposite  equality, 
with  the  sine  of  an  angle  of  the  first  quadrant,  between  0° 
and  90°>  inclusive. 


§64]  THE   SINE   FAMILY.  99 

To  find  such  angle  for  any  given  angle,  note : 

(a)  The  sine  of  a  negative  angle  is  the  negative  of  that  of 
the  same  angle  taken  positive  : 

sin  (-  A0)  =  -  sin^l0. 

(6)  All  multiples  of  360°  may  be  dropped.  Call  the 
remainder  R. 

(c)  If  R  is  less  than  90°,  it  is  the  required  angle. 

(d)  If  R  is  >  90°  and  <  180°,  take  180°  -  R. 
0?)    If  R  is  >  180°  and  <  270°,  take  R  -  180°. 
(/)  If  R  is  >  270°  and  <  360°,  take  360°  -  R. 
(#)  And  then  if  A°  is  the  given  angle, 

(i)        sin  (180  -  R)   =  sin  A°,  case  (<*)• 

(ii)       sin  (72  —  180°)  =  —  sin  J.°,  case  (e). 

(iii)      sin  (360  —  R)    =  —  sin  A°,  case  (/). 

In  practice  it  is  best  to  sketch  the  terminal  of  the  given  angle. 

(Ji)  If  the  terminal  is  in  the  first  quadrant,  its  principal 
angle  is  the  required  angle. 

(i)  If  the  terminal  is  in  the  second  quadrant,  draw  the 
terminal  symmetric  to  the  vertical  and  take  its  principal  angle. 

(/)  If  the  terminal  is  in  the  third  quadrant,  draw  the  oppo- 
site terminal  and  take  its  principal  angle  (opposite  equality). 

(&)  If  the  terminal  is  in  the  fourth  quadrant,  draw  the 
terminal  symmetric  to  the  horizontal  and  take  its  principal 
angle  (opposite  equality).* 

EXERCISE. 

Give  the  angles  less  than  90°  and  greater  than  0°,  which  have  sines 
equal,  in  numerical  magnitude,  to  those  of  the  following  angles,  stating 
in  each  case  the  proper  sign,  and  using  a  diagram : 

136°,  172°,  185°,  200°,  275°,  246°  34' 53",  301°  23',  -  30°  10',  100°  45', 
-185°  54',  -13°  13' 13",  400°,  -  400°,  500°,  -500°,  1200°,  -1290°, 
1800°,  -  1800°,  2500°,  -  2500°. 


*  Some  tables  are  so  arranged  that  the  sine  of  any  angle  less  than  360° 
can  be  taken  from  the  table  directly.     See  Hussey's  Tables. 


100  PLANE   TRIGONOMETRY.  [§65 

§  65.     Supplementary  Angles. 

Any  two  angles  whose  sum  in  degree  measure  is  180°,  or, 
in  radian  measure,  ir  radians,  are  said  to  be  supplementary. 
Each  is  called  the  supplement  of  the  other. 

They  have  the  forms  180°  -  A°,  A° ;  ir  -  0,  0.  To  con- 
struct 180°  —  A%  lay  out  from  the  left-hand  horizontal  line 
an  angle  equal  to  the  angle  A°,  reversed  in  sign.  Thus 
the  terminal  of  an  angle  and  that  of  its  supplement  are 
symmetrically  inclined  to  the  vertical. 

.-.  sin  (180°  -  A°)  =  sin  A° ;  sin  (ir  -  0)  =  sin  0, 

or,  the  sine  of  an  angle  is  the  sine  of  its  supplement. 

EXERCISES. 

1.  Show  from  a  diagram  that  sin  (180°  —  A0)  =  sin  A°. 

2.  If  the  sines  of  two  angles  are  equal,  are  the  angles  necessarily 
supplementary  ? 

3.  Does  it  follow  that  because  the  terminals  of  supplementary  angles 
are  symmetric  to  the  vertical,  all  angles  whose  terminals  are  symmetri- 
cal to  the  vertical  are  supplementary  ? 

4.  Name  five  pairs  of  supplementary  angles  which  have  the  same  pair 
of  terminals.  How  many  pairs  of  supplementary  angles  have  the  same 
pair  of  terminals?  If  A  is  an  angle  of  a  given  terminal,  give  the  gen- 
eral formulas  for  all  supplementary  pairs  having  this  terminal. 

5.  Select  five  angles  of  the  first  quadrant  and  determine  their  sup- 
plements, illustrating  the  positions  of  the  terminals  by  diagrams.  Do 
the  same  for  five  angles  in  each  of  the  other  quadrants. 

6.  Do  the  same  for  five  negative  angles  of  each  quadrant. 

7.  If  A,  B,  C,  are  the  angles  of  a  triangle,  give  three  formulas  con- 
necting these  angles  by  sines. 

8.  Connect  by  sines  pairs  of  opposite  angles  of  a  parallelogram. 

9.  Connect  by  sines  the  adjacent  angles,  when  one  straight  line  meets 
another.  Also  the  exterior  angle  of  a  triangle  with  the  sum  of  the  inte- 
rior opposite  angles. 

10.  Select  five  angles  in  circular  measure  and  determine  their  sup- 
plements. 

11.  What  is  the  supplement  of  45°  -  A°1  Of  |  +  0? 


THE   SINE  FAM'LY.  J(jj 

REVIEW   EXERCISES. 

1.  If  45°  is  a  special  solution  of  the  equation  sin  A  =  — - — ,  what  is 
the  general  solution ?  +  v2 

2.  How  are  the  terminals  of  180°  -  18°,  180°  +  18°,  360°  -  18°, 
related  to  that  of  18°?  How  are  the  sines  of  these  angles  related  to 
that  of  18°? 

3.  How  are  the  terminals  of  all  angles  having  the  same  sine  as  30°, 
related  to  the  terminal  of  30°? 

4.  How  are  the  terminals  of  all  angles  having  the  opposite  sine  to 
that  of  60°,  related  to  the  terminal  of  60°? 

5.  How  is  the  terminal  of  the  supplement  of  an  angle  related  to  that 
of  the  angle  ? 

6.  What  is  the  supplement  of  120°?  Of  -120°?  Of  360°?  Of  -800°? 

7.  What  is  the  supplement  of  -?  Of  -  -?  Of  4tt?  Of  -  8tt? 

3  3 

8.  If  the  sine  of  A  =  $,  what  is  the  sine  of  the  supplement  of  A  ? 

9.  State  how  the  terminals  of  the  following  angles  are  related  to  the 
border  lines  of  the  quadrants,  and  give  the  sign  of  the  sine : 

±45°;    ±60°;     ±f;   ±£;    ±5*;    ±?£;    ±180°;    ±135°;    ±120°; 
o         12  o  o 

±240°;   ±225°;  ±330°;  ±315°;  ±2;  ±^£;  ±  2n7r  +  f ;  ±(2  n  +  1) 


9. 


2^2'^  4 


r-=S  ±<2n-l)»+^. 

10.  State  with  reference  to  each  of  the  following  angles  the  quadrant 
to  which  it  belongs,  the  sign  of  its  sine,  and  the  angle  of  the  first  quad- 
rant whose  sine  is  numerically  the  same : 

150°;   120°;    240°;    225°;   330°;    105°;    165°;    195°;   255°;   480°;  498° 

510°;  585°;  555°;  975°;  1305°;  1590°;  1665°;  ^;  !^;  ^;  il^;  11^ 

2mr  +  ^;  (2n  +  l),r-|;  (2„-l)»  + JL;  2»r-|j  (2n-l)»+|| 

(2n  +  l)T+-;    -405°;   -390°;    -420°;  -840°;    -1200°;   -1305° 
-1020°;  -1665°;  -7|tt;   -  5fv;  -8fir;  -8|*j  -6£tt;  -IS}}*. 

11.  Express  in  radians  46°  30' ;  £  of  a  right  angle ;  49°  43'  45"  ;  1'. 

12.  Express  the  angles  in  Ex.  11  in  terms  of  it  radians. 

Q  if, 

13.  Express  in  degrees,  minutes,  and  seconds  :  -^- ;  2  7rr ;  .8r ;  3.1416r ; 

14.  Taking  tt  as  V>  now  ^ar  °^  *s  a  1-foot  line  when  it  subtends  an 
angle  of  1"  ? 


192 


PLANE   TRIGONOMETRY. 


[§66 


15.  What  angle  at  the  earth's  centre  does  the  moon  subtend,  supposing 
her  diameter  4000  miles  and  her  distance  240,000  miles,  taking  it  =  3.1  ? 

16.  How  far  off  is  a  bright  object  350  feet  tall  when  just  visible? 
(2 1'  is  the  limit  of  average  eye-vision  for  a  bright  object  on  a  black 
background.) 

17.  Find  two  angles  whose  difference  is  1°  and  whose  sum  is  lr. 

18.  Divide  77°  into  two  parts  such  that  the  number  of  English  sec- 
onds in  one  is  the  number  of  French  seconds  in  the  other. 

19.  Name  the  quadrants  to  which  the  following  points  belong  :  (2, 4)  ; 
(-3,  4);  (3,  -4);  (-3,  -4);  (2,  30°);  (-2,  30°);  (2,  -30°); 
(-2,  -30°). 

20.  Define  the  sine  of  an  angle. 

21.  If  modulus  is  27.8  and  sine  =  0.345,  what  is  the  ordinate  ? 

22.  If  the  ordinate  is  3.18  and  the  sine  is  0.432,  what  is  the  modulus? 

23.  If  the  modulus  is  31.4  and  the  ordinate  is  ±  16.3,  what  are  the 
sines,  to  three  places  ? 

24.  A  tree  is  50  feet  high  and  subtends  an  angle  whose  sine  is  0.37 
from  a  point  in  the  horizontal  plane  of  its  ground-line ;  how  far  is  it  from 
that  point  to  the  top  of  the  tree  ? 

§  66.     Construction  of  the  Terminals  for  Given  Values  of  the  Sine. 
Required  the  terminals  when  sin  0  =  J .     Consider  9  as 


modulus  and  5  as  ordinate. 


Draw  a  circle  of  radius  9  about 
the  origin.  Lay  out  on  the 
axis  of  ordinates  a  distance  OD 
equal  to  +  5.  Then  draw  NDM 
parallel  to  the  initial  line,  cut- 
ting the  circle  in  iV,  M.  The 
required  positions  of  the  ter- 
minals are  OM,  ON.  If  0  is 
some  one  of  the  angles  locat- 
ing the  terminal  OM,  as,  for 
example,  the  principal  angle 
as  in  the  diagram,  then, 


Fig.  30. 

2w7T  +  e 

(2n  +  l)w  -  61 
are  the  complete  set  of  angles  whose  sine  is  f  (§  62) 


,  or,  mir  +  (-  l)m0, 


§67] 


THE   SINE   FAMILY. 


103 


If  the  sine  had  been  given  as  —  f,  the  only  change  in  the 
preceding  construction  would  have  been  to  lay  out  OB  5  units 
downward  from  0,  instead  of  upward. 

In  constructing  terminals  for  table-sines,  cut  the  table- 
value  down  to  the  nearest  tenth  or  nearest  hundredth. 

EXERCISES. 

1.  Construct  the  double  terminals  for  each  of  the  following  values 
taken  as  sines  (using  coordinate  paper)  : 

-  f ,  i,  - 1, 1,  -i,h-  h  h  -  h 

Are  any  of  the  preceding  values  impossible  ? 

2.  Selecting  a  line  to  represent  unity,  construct  a  line  to  represent 
+  V2  ;  a  line  for  V3 ;  for  V5.    Can  you  see  from  the  construction  for  V2, 

what  angles  have  ±i  for  sine?    What—?    What  ^?    What -=-^?? 

V2.  V2  2  2 

3.  Where  is  the  terminal  when  the  sine  isl?  — 1?  0?  What  are 
the  general  expressions  for  the  corresponding  angles  ? 

4.  Construct  terminals  for  five  sines  in  the  table  of  sines,  and  meas- 
ure the  angles.     Compare  with  table-values. 


§  67.     Line  Picture  of  the  Sine. 

ordinate      MP     MtPt    .-   .,  ,   . 

Since    sine  = n —  =  -^7  =     '     ,   it   the   modulus   is 

modulus      OP       OPx 

taken  as  unity,  the   sine   becomes,  on   the   same   scale,  the 
ordinate.     If,  therefore,  0PX 
(Fig.  31)  is  taken  as  a  unit, 
M1P1  is  the  sine,  or 

sin  6  =  ordinate  M1PV  (A). 
(A)  is  frequently  spoken  of 
as  the  line  definition  of  the 
sine. 

Not  many  years  ago  all  text- 
books on  trigonometry  used  this 
definition  exclusively.  Two  objec- 
tions have  led  to  its  abandonment.  Fig.  31. 
(1)  Its  use  gives  a  false  impres- 
sion at  the  outset,  namely,  that  the  sine  is  a  line,  and  a  line  of  a  cer- 
tain length,  that  of   the  diagram  in  the  text-book  studied.     The  fact 


-,£.1                 ^_       .      1 

7                                          \                    jt^ 

'                                             .     \  \l      ^ 

7                               4^ 

n-                                                  At 

I                         +'    %+ 

t                 *'      l\? 

i£  .*           u 

_£^               tr 

_l                                                  _,*/         .if. 

\                                                     1/ 

v                                                   / 

\                                               / 

\                                         / 

s^                                                          / 

^s                               ^-' 

■---:=.              .-'' 

104 


PLANE   TRIGONOMETRY. 


[§68 


is  that  the  sine  is  not  a  line,  but  a  ratio,  as  already  seen.  The  diagram 
means  merely  that  the  line  MP  represents  the  sine  on  the  same  scale 
(any  scale)  on  which  OP  represents  unity.  Since  the  unit-representative 
is  arbitrary  in  length,  so  is  the  line  representing  the  sine.  If  the  line 
representing  a  unit  is  taken  half  as  long  as  that  above,  the  sine-line  is 
reduced  one-half.  (2)  The  second  objection  to  the  use  of  the  line  defi- 
nition is  that  those  who  learn  trigonometry  with  unity  as  the  modulus, 
find  trouble  in  introducing  any  other  modulus,  whereas,  with  the  general 
modulus,  r,  used  from  the  outset,  no  difficulty  is  met  in  assigning  to  r  the 
special  value,  1. 

The  ratio  definition  was  first  used  in  America  at  the  University  of 
Virginia,  by  Professor  Bonny  castle. 

The  line  definition  continued  to  be  used  in  American  text-books  to  a 
very  recent  day,  and  is  still  in  very  general  use  in  engineering  field 
books.  Instead  of  abandoning  completely  the  line  definition,  as  is  now 
the  drift  in  text-books,  it  appears  to  the  author  advisable  to  hold  it  for 
such  use  as  is  made  in  §  69.  It  presents  to  the  eye  certain  properties, 
and  thus  teaches  these  properties  with  greater  lucidity  than  does  the  ratio 
definition. 


§  68.    The  Expression  "Sine  of  the  Arc." 

It   has    been    shown   (§  47,  c,  4)  that    when    the  radius 
is    taken    as    unity,    the    number    representing    the    arc    is 

the  same  as  that  representing 
the  corresponding  central  angle 
in  radian  measure.  Conse- 
quently, when  the  line  defini- 
tion was  in  use,  the  sine  of 
the  angle,  as  now  used,  was 
called  the  sine  of  the  arc.  The 
notation  for  an  angle  whose  sine 
is  any  special  value,  as  |,  is 
sin-1!-  This  is  generally  read, 
"the  angle  whose  sine  is  §," 
frequently,  "  the  arc  whose  sine 
is  f."  This  is  a  lingering  influence  of  the  line  definitions, 
as  is  the  designation  used  in  German  text-books,  "  arcns 
sinus."  The  relative  magnitude  of  arc  6  and  ordinate  MXP^ 
(Fig.  32)  expresses  graphically  the  size  of  an  angle  com- 
pared with  its  sine. 


.-- •«. 

"2Z.                       'a. 

7                                Sit 

7                                                  '&* 

r                                  <*~X 

2                             s>  -W 

t                          .<"         rt 

L                      c.    ^              i  \ 

Js          .hi 

-                        hr~ 

f                                           Jii; 

\                                              / 

^                                         -^ 

-f 

-    S                           _z 

J7 

Fig.  32. 


§69c] 


THE   SINE   FAMILY. 


105 


§  69.    Line  Picture  of  the  Sines  of  all  Angles,  and  its  Story. 


\ 


x 


J* 


: 


Fig.  33. 


Draw  the  unit-circle  about  0.  It  cuts  the  terminal  OP 
at  Px  (Fig.  33).  Then  M1P1  is  the  sine  of  6.  The  ordinate 
at  any  point  represents  the  sine  of  all  angles  whose  terminals 
pass  through  that  point.  Thus  the  ordinates  of  the  unit-circle 
make  up  the  sum  total  of  all 
possible  values  of  the  sines  of 
all  angles. 

From  the  diagram  of  the 
line  values  of  sines  (Fig.  33) 
there  can  be  learned  more 
readily  than  in  any  other  way  — 
because  seen  by  the  eye  —  cer- 
tain very  important  properties 
of  the  sine. 

(a)  When  the  terminal  is 
close  to  the  initial  line,  the  sine 
is  small. 

(6)  When  the  terminal  is  on  or  opposite  the  initial  line, 
the  sine  is  zero. 

(<?)  Starting  the  terminal  from  coincidence  with  the  initial 
line,  and  letting  it  move  counter-clockwise  around  the  origin, 
the  sine  sets  out  with  the  value  zero,  and  increases  as  do 
the  numbers  of  a  continuum,  until  the  terminal  reaches  the 
upright  vertical,  when  the  sine  becomes  the  radius,  or  1. 
As  the  terminal  passes  the  position  of  the  upright  vertical, 
the  sine  begins  to  decrease,  and,  as  the  terminal  moves  on, 
the  sine  continues  decreasing  as  do  the  numbers  of  a  con- 
tinuum, until  it  reaches  the  value  0,  when  the  terminal  co- 
incides with  the  left-hand  horizontal.  The  sine  continues  to 
decrease  as  do  the  numbers  of  a  continuum,  until  the  terminal 
is  vertical,  downward,  when  it  is  —  1.  It  then  begins  to 
increase,  and  increases  as  do  the  numbers  of  a  continuum 
until  the  terminal  is  again  in  coincidence  with  the  initial  line, 
when  it  returns  to  the  value  with  which  it  started  out,  zero. 


106  PLANE   TRIGONOMETRY.  [§  69  d 

As  the  terminal  moves  again  over  its  circuit  the  sines  go 
again  through  the  same  order  of  values  ;  and  so  on,  over  and 
over  again,  as  the  terminal  continues  to  turn. 

(d)  With  a  clockwise  turn  of  the  terminal,  the  same  thing 
takes  place  in  the  reverse  order.  (The  student  may  state 
verbally  what  happens.)     Thus  : 

(e)  The  sine  is  never  greater,  numerically,  than  1. 

(/)  It  can  take  all  numerical  values  lying  between  +  1 
and  —  1,  inclusive. 

(</)  It  is  zero  whenever  the  terminal  is  on  or  opposite  the 
initial  line. 

Qi)  It  is  +1  whenever  the  terminal  is  the  upright  vertical, 
and  —  1  whenever  the  terminal  is  the  downright  vertical. 

(%)  When  the  sine  is  zero,  the  terminal  is  on  or  opposite 
the  initial  line,  and,  therefore,  the  angle  is  an  even  or  odd 
number  of  half  turns,  or 

2  mr,   or   (2  n  +  l)7r, 

or,  2  n  -180°,   or   (2w  + 1)180°. 

Important  special  cases  are  : 

sin0°=0,   sin  (±  180°)  =  0,   sin  (  ±  360°)  =  0, 
sin  (  ±  tt)  =  0,   sin  (  ±  2  tt)  =  0, 

where  the  general  formula  is 

sin  mir  =  0,  or  sin  m  •  180°  =  0, 

where  m  is  any  positive  or  negative  integer. 

(j)  When  the  sine  is  +1,  the  terminal  is  the  upward 
vertical  and  the  angle  is  of  the  form 

2*wr  +  £,  or  2 n  •  180°  +  90°, 

A 

or,  (4  n  +  1)  i  or  (4  n  +  1)  90°. 

A 

Therefore,  in  general, 

sin(4w+l)£=l,  or  sin  (4  n  +  1)  90°  =  1. 

8 


§69/]  THE   SINE   FAMILY.  107 

And  in  particular, 

sin  —  aas  1,  sin  5  •  —  =  1,  etc. 

—  A 

sin  90°  =  1,  sin  5  x  90°  =  1,  etc. 

Sin(~8|)  =  1,  sin(-7|)  =  l,etc. 
sin(-  3  x  90°)=  1,  sin(-  7  x  90°)=  1,  etc. 

(6)  When  the  sine  is  —  1,  the  terminal  is  the  downward 
vertical,  therefore, 

sin(-  90°)=  -  1,  sin(-  5  x  90°)=  -  1,  etc. 

sin(~l)  =  _1,  sin("5*i)=  -1'etc-> 

with  the  general  statement, 

sin(4n-l)90°  =  -l, 

or,  sin(4n-l)£  =  -l, 

8 

where,  as  in  all  formulas  of  this  kind,  here  as  elsewhere,  n  is 
any  positive  or  negative  integer. 

(7)  The  sine  of  an  angle  is  less  than  the  corresponding 
arc  on  the  unit-circle,  that  is,  the  arc  of  the  principal  angle. 
This  is  evident  at  once  on  the 

unit-circle,  for  the  arc  PAPX 
(Fig.  34)  is  longer  than  the 
straight  line  PPX ;  therefore, 
the  half  of  arc  PAPX  is  longer 
than  the  half  of  the  straight 
line  PPV 

1  Fig.  '34. 

or,  MP  <  arc  AP, 

or,  sin  0  <  arc  AP. 

It  has  been  mentioned  in  §  68  that  when  the  radius  is  taken 
as  unity,  the  number  representing  the  arc  on  the  unit-circle 
is  the  same  as  the  number  representing  the  corresponding 
central  angle.  Therefore  an  equivalent  statement  to  that 
above  is, 

The  sine  of  an  angle  is  less  than  its  radian  measure. 


108  PLANE   TRIGONOMETRY.  [§  69 1 

The  same  proposition  is  also  readily  proven  for  any  other 
radius  than  unity.     For  in  Fig.  34, 

straight  line  GrBQ  <  arc  Q-SQ, 

.:  ordinate  BQ  <  arc  SQ, 

ordinate  BQ        arc  SQ 
modulus  OQ       radius  OQ^ 

or,  sin  6  <  radian  measure, 

or,  sin  6  <9. 

(m)  A  slight  change  in  the  angle  makes  a  slight  change 
in  the  sine. 

For  the  angle  A  OP1  (Fig.  35),  the  sine  is  MXPV     For  the 
larger  angle  A  OPv  the  sine  is  M2P2. 

.\  sin  AOP2- sin  A  OPt 

Evidently,  then,  with  diminishing 
difference  in  the  angles,  SP2  di- 
minishes, becoming  zero  when  the 
angle  difference  is  zero. 

When  the  angle  difference  be- 
comes less  than  any  assignable  quantity,  so  also  does  the 
sine  difference.  An  equivalent  statement  to  this,  and  the 
one  most  commonly  used  is : 

The  sine  varies  continuously  as  the  angle  varies  continuously. 
This  is  also  differently  expressed  by  saying  that  if  angle 
values  form  a  continuum  from  A  to  i?,  the  sine  values  form 
a  continuum  from  the  sine  of  A  to  the  sine  of  B.  For  ex- 
ample, as  the  angle  passes  from  zero  to  90°,  the  sine  passes 
through  all  those  values  indicated  as  ordinates  on  the  unit- 
circle  from  0  to  1.  Similarly,  as  the  angle  changes  continu- 
ously from  45°  to  135°,  the  sine  passes  through  all  values 

V2  V2 

from  •— -  to  1,  and  then  from  1  to  -—  again.     It  waxes  in 

length  as  does  a  stretching  rubber  string,  and  wanes  in  length 
as  does  a  contracting  rubber  string.  The  illustration  ex- 
presses fairly  well  what  is  meant  by  a  continuous  change, 


§69n]  THE   SINE   FAMILY.  109 

a  change  without  sudden  jumps  in  value,  without  omitting 
intermediate  values  of  a  continuum.  One  can  easily,  how- 
ever, get  a  false  impression  from 
the  illustration,  if  any  idea  of 
the  rate  of  change  along  the  con- 
tinuum is  allowed  to  enter  the 
mind.  Suppose,  for  instance, 
one  considers  the  ordinates  of  the 
straight  line  AB  (Fig.  36)  from 
A  to  B.  They  form  a  continuum. 
The  ordinates  of  the  circle  APB 
form  exactly  the  same  continuum, 

as  do  also  the  ordinates  of  any  other  curve,  like  A  QB,  which 
has  no  ordinate  between  A  and  B  higher  than  OB-  This  is 
seen  by  drawing  a  line  RQP  parallel  to  the  initial  line.  This 
gives  three  equal  ordinates,  so  that  for  any  ordinate  of  any 
one  of  the  curves  (line),  there  is  an  equal  ordinate  on  the 
others.  Thus  the  sets  of  ordinates  form  the  same  continuum, 
whether  from  the  straight  line  or  from  the  circle.  Evi- 
dently the  rate  of  rise  of  the  ordinates  of  the  circle  is  not 
uniform.  The  rise  is  rapid  at  the  outset  at  A,  and  is  a  di- 
minishing rate  of  rise  as  the  terminal  turns  uniformly  from 
the  position  OA  to  OB. 

As  much  of  the  notion  of  continuity  as  we  desire  at  pres- 
ent to  give,  is  covered  by  the  statement : 

If  two  angles  in  circular  measure,  or,  what  is  the  same 
thing,  two  arcs  on  the  unit-circle,  differ  by  less  than  any 
assignable  magnitude,  so  will  their  sines. 

This  follows  from  the  diagram  (Fig.  35),  for  the  straight 
line  SP2  is  shorter  that  the  curve  P^Py  that  is,  the  sine 
difference  is  always  less  than  the  arc  difference. 

(n)  For  a  given  small  change  in  the  angle,  the  change  in 
the  sine  is  smaller  the  nearer  the  terminal  of  the  angle  is  to 
the  vertical ;  and  the  larger,  the  nearer  the  terminal  is  to  the 
horizontal.  In  particular,  if  two  angles  are  very  near  90°, 
the  difference  in  their  sines  is  very  small  compared  with  the 
difference  in  the  angles ;  whereas,  if  the  two  angles  are  very 


110 


PLANE   TRIGONOMETRY. 


[§69o 


near  zero,  the  difference  in  the  sines  is  very  nearly  equiva- 
lent to  the  difference  in  the  angles  in  radian  measure.  The 
student  is  expected  to  see  this  only  in  so  far  as  it  is  clear 
from  the  diagram,  that  is,  see  it  with  the  eye.  Later  the 
same  thing  will  be  investigated  by  the  aid  of  formulas. 
Often  we  see  on  the  diagram  what  a  formula  never  makes 
any  clearer. 

(0)  When  an  angle  is  small,   doubling   the   angle   very 
nearly  doubles  the  sine. 

Motion  along  the  arc  at  4$  on  the  cir- 
cle, is  very  nearly  vertical,  and  the  in- 
crease in  the  ordinate  is  very  nearly  the 
same  as  in  the  arc. 

(jt>)  However,  sin  2  a  is  never  2  sin  a, 
except  when  a  is  zero  or  some  multiple 
of  ir. 

For,  let  APB  (Fig.   38)   be   a   semi- 


FlG.  37. 


IT 


circle,  with  a  >  0  and  <  — •     Then, 


sin  2  a 


MP 


and 


sm  a  = 


Fig.  38. 


OP' 

In  the  triangle  BOP, 
BP<B0+  OP,  i.e.  BP<%  OP. 

.       ^MP        0   .        ^MP 

'   sm  a  >  -  ^^,  or  2  sm  a  >  — — , 


MP 
BP' 


2  OP 

or,  2  sin  a  >  sin  2  a,  numerically. 

This  is  true  for  all  values  of  a.  See  §  139  (1).  We 
introduce  the  word  "  numerically "  above,  to  correspond 
with  the  result  which  would  be  obtained  if  the  diagram  were 
turned  over  about  BA  as  an  axis. 

We  give  the  statement,  sin  2  a  is  not  2  sin  a  here,  because 
the  desire  to  treat  2  as  a  factor  with  the  word  "  sin  "  mani- 
fests itself  quite  promptly  among  average  students. 


EXERCISE. 


Deduce  all  the  propositions  from  (a)  to  (jo),  using  the  ratio  definition 
for  the  sine,  and  without  making  use  of  the  unit-circle. 


§70]  THE   SINE  FAMILY.  Ill 

LABORATORY  EXERCISES. 

1.  Draw  a  circle  of  one  foot  radius.  Divide  it  into  five-degree  spaces  ; 
measure  the  ordinates  in  decimals  of  a  foot,  and  compare  with  a  table  of 
sines,  and  show  that  the  sines  for  the  second,  third,  and  fourth  quadrants 
can  be  expressed  in  terms  of  those  of  the  first  quadrant. 

2.  On  the  same  circle  measure  the  sine  of  10°  and  the  sine  of  5°,  and 
see  that  sin  10°  <  2  sin  5°.     Test  several  other  double  angles. 

3.  Show  from  the  diagram  of  Ex.  1  that  sin  15°  <  3  sin  5°. 

4.  Show  from  the  diagram  of  Ex.  1  that  sin  20°  <  4  sin  5°. 

§  70.   The  Sine  as  a  Function  of  the  Angle. 

Whenever  a  quantity  y  is  so  related  to  another  quantity  x, 
that  y  can  be  calculated  when  x  is  given,  then  y  is  said  to 
be  a  function  of  x.  Calculations  involve  four  operations, 
—  addition,  subtraction,  multiplication,  division.  In  fact, 
these  are  the  only  operations  involved  in  calculations.  When 
y  is  reached  by  a  finite  number  of  calculation  operations  on 
sb,  y  is  said  to  be  an  algebraic  function  of  x.     Examples  are  : 

y  =  2x, 
y=2x±3, 

f  8  a*+  6  a*  -  8  * 


V 


etc. 


9s5_6z2  +  3 

Here  y  is  in  each  case  derived  from  #  by  a  finite  number  of 
calculation  operations.  Roots  of  algebraic  functions,  where 
the  root-indices  are  integers  or  common  fractions,  are  also 
reckoned  as  algebraic  functions,  even  though  it  may  not  be 
possible  to  calculate  the  function  exactly. 

Functions  that  are  not  algebraic  are  said  to  be  transcen- 
dental, since  their  calculation  implies  an  infinite  number  of  one 
or  more  of  the  fundamental  mathematical  operations.  Such 
functions  can  evidently  be  calculated  only  approximately, 
this  being,  however,  as  indicated  above,  no  test  of  the  trans- 
cendental character  of  a  function. 


112  PLANE   TRIGONOMETRY.  [§  70 

That  the  sine  of  an  angle  is  a  function  of  the  angle,  in  the 
sense  of  the  foregoing  definition  of  a  function,  is  evident  at 

once  from  the  definition  of  a  sine  ( ).     For  one  can 

Vmodulus/ 

draw  the  ordinate,  measure  it  and  the  corresponding  modulus, 
divide,  measure  the  angle.  It  is  evident,  however,  that  while 
such  a  procedure  establishes  the  fact  that  the  sine  is  a  function 
of  the  angle,  the  value  of  the  sines  of  all  angles  could  not  be  ob- 
tained with  any  great  degree  of  accuracy  from  this  procedure 
without  great  care  in  the  measurements.  The  ratio  definition 
of  the  sine  is  not  given  with  a  view  to  its  use  in  calculating 
the  tabulated  sines  of  angles,  but  rather  with  the  object  of 
calculating  the  ordinate  when  the  angle  and  modulus  are 
given,  or  calculating  the  modulus  when  the  angle  and  ordi- 
nate are  given.  That  is,  the  definition  is  not  used  to  calculate 
a  table  of  sines  of  angles.  Frequently,  however,  from  meas- 
ured ordinates  and  moduli,  sines  are  calculated  and  the 
corresponding  angle  determined  by  comparison  of  the  result- 
ing calculated  sine  with  a  table  of  sines.  It  will  be  shown 
in  §  156  that  if  6  is  the  radian  measure  of  an  angle, 

sin 6  =  6  —  —  +  —  —  —  -f  —  ...  etc.,  without  end. 
[3     [6.    (I     [9 

From  this  relation  the  sine  of  any  angle  can  be  calculated 
to  any  desired  degree  of  accuracy,  but  never  when  6  is  other 
than  zero  can  it  be  determined  completely  from  this  series, 
since  an  infinite  number  of  calculation  operations  is  indicated. 
The  sine  is  thus  a  transcendental  function  of  the  angle,  and 
in  most  cases  can  be  known  only  approximately. 

While  the  formula  above  for  the  sine  would  be  used  to 
calculate  the  sine  of  any  given  angle,  if  that  were  unknown,  it 
is  not  the  formula  from  which  tables  of  sines  of  angles  have 
been  calculated.  It  will  be  shown  later  that  when  the  sine 
of  any  angle  is  known,  that  of  an  angle  differing  from  it 
slightly  can  be  determined  more  readily  than  by  using  this 
formula.     The  chief  point  we  wish  to  make  here  is,  that  the 


§72]  THE   SINE   FAMILY.  113 

sine  is  a  function  of  the  angle ;  that  is,  that  when  the  angle 
is  given,  the  sine  can  be  calculated.  These  calculations  have 
been  made  with  great  care  for  all  angles  between  0°  and  90°, 
at  intervals  of  every  "  second  of  arc  "  for  small  angles ;  for 
every  10"  for  larger,  but  still  small,  angles ;  then  for  every  1', 
or  for  every  10',  the  angle -interval  depending  upon  the  place 
of  the  table. 

Such  tables  of  sines  are  frequently  called  tables  of  natural 
sines,  to  distinguish  them  from  the  tables  of  logarithms  of  the 
same  numbers,  called  tables  of  logarithmic  sines. 

§  71.   Many- valued  Functions. 

When  to  each  value  of  the  quantity  x  corresponds  a  single 
value  of  the  calculated  quantity  y,  y  is  said  to  be  a  single- 
valued  function,  or  unique  function,  of  the  quantity  x. 
Examples  are  the  algebraic  functions,  free  from  radicals, 
previously  given.  If  y  =  V#2  -f-  3x  —  7,  to  each  value  of  x 
will  correspond  two  values  of  y.  In  this  case,  and  all  similar 
ones,  y  is  called  a  two-valued  function  of  x.  While,  as 
already  seen,  the  sine  is  a  single-valued  function  of  the  angle, 
the  angle  is  a  many-valued  function  of  the  sine,  since  all 
angles  with  the  same  terminal,  or  with  terminals  symmetric 
to  the  vertical,  have  the  same  sine  (§  62). 

§  72.    The  Sine  as  a  Periodic  Function. 

When  it  is  desired  to  indicate  that  y  is  a  function  of  x 
without  stating  the  character  of  the  relation  specifically,  we 
write  y  =  F(x). 

This  is  read,  "y  equals  a  function  of  #."  If  two  or  more 
functions  are  to  be  used  at  the  same  time,  some  other  letter 
may  take  the  place  of  F.     Thus  we  might  have 

y  =  F(x),   y  =  0(x),   y  =  <j>(x),   y=G(x),  etc. 

In  such  cases,  if  the  functions  are  read  in  succession,  dis- 
tinction is  made  among  them  by  reading  the  function  letter, 
as  "y  equals  the  F-i unction  of  as,"  uy  equals  the  theta- 
function  of  #,"  etc. 


114  PLANE   TRIGONOMETRY.  [§72 

If  y  =  F(x),  then  yt  =  F^x^  means  that  yx  is  the  special 
value  which  y  takes  when  x  is  given  the  special  value  xv 
For  example,  if 

y  =  z2  -f-  3  z  -  7  =  F(x), 

F(x1)  =  x12  +  Sx1-1, 

and  F(2)  =  22 +  3-2-7  =  3,  etc. 

Similarly,  if  y  =  FQc),  then  yx  =  F(xx  +  K)  means  that  y1 
is  the  special  value  which  y  takes  when  x  takes  the  special 
value  x1  +  h. 

With  these  agreements,  if  y  =  F(x),  then 

F(xJ  =  F(xx  +  K) 

signifies  that  y  takes  the  same  value  for  x  =  x1  as  for 
x  =  a?j  +  h,  and 

F(x{)  =  JX^  4-  K)  =  .FOi  +  2  K)  =  F(xx  +  3  K)  =  F(xx  -f  4  ft),  etc. 
=  #<>!  -  A)  =  #0^  -  2  A)  =  .F(>i  -  3  A),  etc., 

signifies  that  all  values  of  x  separated  from  x1  by  multiples  of 
the  value  ^,  give  the  same  value  to  the  function  y. 
When  we  write 

F(x)  =  F(x  +  nh), 

where  n  is  any  integer,  positive  or  negative,  and  x  is  free 
from  subscripts,  it  signifies  that  for  any  value  whatever  of  x, 
values  of  x  separated  from  this  value  by  multiples  of  h  give 
all  the  same  value  to  the  function  as  did  the  initial  value. 
Such  a  function  is  called  a  periodic  function,  and  h  is  called 
the  period. 

We  have  seen  that  for  the  sine, 

sin  (0)  =  sin  (6  +  n  •  2  tt), 

for  all  integral  values  of  n,  and  for  all  values  of  0  (§  62), 
and  that  this  is  the  only  angle  relation  which  leaves  the  sine 
unchanged  both  in  magnitude  and  sign. 

Thus  the  sine  is  a  periodic  function  whose  period  is  2  7r. 


§73]  THE   SINE  FAMILY.  115 

EXERCISE. 

Point  out  the  periodicity  of  the  sine  on  the  unit-circle,  taking  the  arc, 
its  number  being  the  same  as  that  of  the  corresponding  angle,  to  repre- 
sent x,  while  the  sine,  or  ordinate,  is  y. 

§  73.    Inverse  Functions.     Notation  for  Angles  having  a 
Given  Sine. 

When  y  is  expressed  directly  in  terms  of  S,  y  is  said  to  be 
an  explicit,  or  direct,  function  of  #,  as  in  y  =  x2  +  2  x  —  1 
or  y  =  sin  x.  When  x  is  given  here,  y  can  be  calculated 
directly.  Generally,  if  y  is  a  function  of  x,  x  is  also  a  func- 
tion of  y  ;  that  is,  if  y  is  given,  x  can  be  calculated.  In  the 
first  of  the  relations  above,  if  x  is  2,  y  is  7.  The  inverse 
process  of  finding  what  x  is  when  y  is  given  as  7,  is  quite 
different.     For  when 

z2  +  2z-l  =  7, 

by  the  usual  process  for  solving  quadratics, 


-2±V4  +  32 
x  = ^— — ! =  2  or  —  4. 

A 

Here,  while  y  was  single -valued,  x  is  two-valued. 
Similarly,  if  y  is  a  cubic  in  x,  as 

y=2x*-5x2  +  ±x-S, 

y  will  be  single-valued,  and  x,  as  shown  in  algebra,  three- 
valued,  with  corresponding  relations  for  quartics,  quintics, 
etc. 

From  the  foregoing  it  may  readily  appear  that  if  one 
quantity,  «/,  is  expressed  directly  in  terms  of  another,  x,  it 
may  be  quite  difficult,  or  impossible,  to  express  x  in  terms  of 
y,  or  to  calculate  x  when  y  is  given.  For  example,  if  y  is  a 
quintic  in  #,  as 

y  =  x5  +  ax*  +  bz?  -h  ex2  +  dx  +  e, 

#  cannot  be  expressed  in  terms  of  y,  and  the  quantities 
a,  5,  <?,  (?,  £,  in  terms  of  radicals. 

A  like  relation  holds  for  algebraic  functions  of  higher 


116  PLANE   TRIGONOMETRY.  [§73 

degree  than  the  fifth,  when  the  coefficients  are  given  the 
general  values,  a,  b,  <?,  etc. 

However,  if  the  quantities  a,  5,  <?,  d,  e,  y,  etc.,  are  numbers, 
x  can  be  calculated  to  any  desired  degree  of  accuracy,  as 
explained  in  algebra  (Horner's  Method). 

We  may  assume,  in  general,  that  if  y  is  a  function  of  x, 
x  is  also  a  function  of  y.  The  one  function  is  said  to  be  the 
inverse  function  of  the  other. 

The  inverse  function  of  any  function  F  is  indicated  by 
F~\     Thus,  if 

y  =  F(x),  (a) 

x  =  F_1(y}.    (Read  :  x  equals  anti-^of  y)    (5) 

These  two  notations  imply  the  same  equation-relation,  or  an 
equivalent  equation-relation,  between  x  and  y.     So  do 

y  =  sin  x,  (1) 

x  =  sin-1  y.  (2) 

As  already  pointed  out,  y  is  here  a  one- valued  function  of  x, 
while  a;  is  a  many-valued  function  of  y.  In  fact  (1)  is  nothing 
but  o         g         7 

ryO  rf«J  ry>i 

The  infinite  degree  of  the  second  member  would  carry 
with  it,  to  one  familiar  with  algebra,  an  infinite  number  of 
values  of  x  for  any  value  of  y.  We  have  already  seen  that 
for  any  given  value  of  the  sine  there  are  an  infinite  number 
of  angles,  differing  by  multiples  of  2  7r,  or  180°. 

The  direct  expression  of  x  in  terms  of  y  from  (3)  we  do 
not  take  up  here.  Read  "  Reversion  of  Series  "  in  any  college 
algebra. 

Relation  (2)  is  read  in  several  different  ways  : 

(i)  x  is  an  angle  whose  sine  is  y  ; 

(ii)  x  is  an  arc  whose  sine  is  y ; 

(iii)  x  is  anti-sine  y  ; 

(iv)  x  is  the  inverse  sine  of  y ; 


§  74]  THE  SINE   FAMILY.  117 

(iii)  is  preferred,  on  account  of  its  brevity.  It  is,  perhaps, 
best  to  use  (i)  until  the  significance  of  the  symbol  is  well 
in  mind. 

The  statements 

j  sin  •  sin-1?/  (3)  )  ,   J  sine  anti-sine  y) 

\  sin-1  •  sin  y  (4)  )    '  {  anti-sine  sine  y ) " 

(3)  is  also  read  :  "  The  sine  of  the  angle  whose  sine  is  y," 
and  thus  makes  immediately  the  mental  impression  that  its 
value  is  none  other  than  y  itself.  After  some  practice,  and 
after  the  notation  is  familiar,  "  sine  anti-sine  "  will  make  the 
same  impression.  Expression  (3)  is  thus  single- valued. 
On  the  contrary,  (4)  may  be  read  "  any  angle  whose  sine  is 
that  of  y."     It  is  thus  many-valued,  being  any  one  of  the 

anSleS  IMT+(-l)-0, 

where  6  is  any  particular  angle  of  the  set  (§  62). 

§  74.    Origin  of  the  Notation,  sin-1. 

a x  is  x.     That  is,  a  •  a'1  •  x  is  x. 

a 

Therefore  a  and  a~\  considered  as  operators  on  #,  annihilate 
the  effect  of  each  other. 

.  Corresponding  to  this  is  the  effect  of  the  expressions  "  sine 
of "  and  "  whose  sine  is,"  as  in  the  expression  "  the  sine  of 
an  angle  whose  sine  is  g,  is  a?." 

Thus,  while 

( Bin  •  sin"1  x  =  x)    ^     correspond  to  each  other, 
(    a  •  a-1  •  x  =  x )  • 

(sin-i-sinz)    (2)     donot. 
(   a-1 • a  -  x   ) 

for  while  the  last  is  x,  the  next  to  the  last  is,  as  already 
pointed  out,  any  angle  whose  sine  is  that  of  x. 

However,  the   origin  of   the  notation,  sin-1  a;,  is   in   the 
similarity  here  noted  in  (1) ;  so  also  loge_1  x. 


118  PLANE   TRIGONOMETRY.  [§74 

EXERCISES. 

1.  Read  in  all  possible  ways  each  of  the  following  expressions : 
sin-?;       **»,       rin-l(-l);       dn-i(-f);       sin--L; 

*,->(_  f)j       sfc-.f;       sin-^;       ^(-1). 

2.  Read  the  following : 

Bin-»i  +  sln-i(-|)i       sin-^-  sin--  i; 

sin->|  +  sin-i(-|);     sin- ^  +  sin-i ( _ ^ .     sin-.|  ±  sin-(-|). 

3.  If  0,  A  are  the  radian  and  degree  measure,  respectively,  for  the 
principal  angle  of  any  expression  of  Ex.  1,  what  is  the  general  expression  ? 
Write  the  general  expressions  for  the  angles  of  Ex.  2,  in  both  radian  and 
degree  measure. 

4.  Read  the  following  expressions  in  all  possible  ways  : 

sin"1-  +  3  sin-1- ;  2  sin"1—  -  5  sin-U ;  2  sin"1!  -  3  sin  — ; 

3  4  _2  _  2  2' 

sin"1^  2  8^-4;     4sin^5-2sin-1(-^>);     siW-§)  -sin"1?; 
3  2  2  \       2  /  V     3/  3 

and  write  the  general  expressions  in  circular  measure  and  in  degree 

measure  for  the  corresponding  angles,  using  the  tables  when  necessary. 

5.  Construct  the  terminals  corresponding  to  each  expression  in  Exs. 
1,  2,  and  4. 

GENERAL  EXERCISES. 

1.  Construct  the  terminals  when  sin  A  has  the  following  values: 

z        4    1.        i  *  -n.    V§.   VI.    -V2.    -V3    1         1 
|,       3,  l,        l,  u,   — ,    — ,    -— -,   -£- ;  -;   --. 

2.  Give,  with  diagrams,  general  solutions  of  the  equations : 

sin  0     =  1 ;  sin  6  =  —  1 ;  sin  6       =  0 ; 

sin  2  $  =  1 ;  sin  3  6  =  -  1 ;  sin  4  0     =  0  ; 

sin  md  =  1 ;  sin  (  —  m0)  =  —  1 ;  sin  2  mO  =  0. 

3.  Name  two  angles  not  in  the  same  quadrant  whose  sines  are  the 
same  as  that  of  17°,  with  diagram  for  terminals. 

4.  Name  two  angles  not  in  the  same  quadrant  whose  sines  are  the 
opposite  of  that  of  17°,  with  diagram  for  terminals. 

5.  What  is  the  best  plan  for  determining  for  any  given  angle  what 
angle  of  the  first  quadrant  has  numerically  the  same  sine  ? 

6.  What  are  the  general  values  of  0  when  6  —  sin-1  1,  $  = 
sin-1  (-1),  6  =  sin-1  (0)?  Of  A  when  SA  =  sin-1  (+  1),  mA  = 
sin-1  (-1),  7  A  =  -  sin-1  (0)  ? 


§75]  THE   SINE   FAMILY.  119 

7.  Write  the  general  form  of  a  periodic  function  of  x  whose  period  is  K. 

8.  Write  the  sine  equation  which  indicates  that  the  sine  is  a  periodic 
function. 

9.  If  sin  3  6  =  15,  may  we  divide  by  3  and  write  sin  6  =  5? 

10.  Give  the  general  formula  for  all  angles  whose  terminals  are 
either  on  the  upright  or  downright  vertical. 

11.  Find  the  number  of  seconds  in  the  angle  0.3r;  the  number  of 
minutes  in  0.4r;  the  number  of  degrees,  miuutes,  and  seconds  in  0.5r  and 
in  A**. 

12.  Express  11"  in  radian  measure. 

13.  The  three  angles  of  a  triangle  have  the  same  numerical  measure, 
one  being  in  degree  measure,  another  in  grade  measure,  the  third  in 
radians;  find  the  third  angle  in  terms  of  7r. 

14.  Construct  the  terminals  of  0,  when  2  0  =  sin-1  § ;  when  3  0  = 
sin-1  (  — £);  when  4  0  =  sin-1  (1);  giving,  in  each  case,  all  possible 
solutions. 

15.  What  angle  of  the  first  quadrant  has  the  same  numerical  sine  as 
2317°? 

16.  How  are  the  sines  of  ir  +  6,  ir  —  6,  2  ir  —  6  related  to  sin  0? 

17.  Express  3^.12  in  radian  measure  in  terms  of  ir. 

18.  When  the  modulus  is  25.4  and  the  sine  —  0.312,  what  is  the 
ordinate  ? 

19.  When  the  ordinate  is  ±  37.2  and  the  sine  is  ±  0.341,  what  is  the 
modulus  ? 

20.  When  the  ordinate  is  ±  21.3  and  the  modulus  is  32.4,  what  are 
the  sines  to  three  figures  ? 

§  75.    Angles  whose   Sines    can   be  determined  readily  from  a 

Diagram. 

We  have  already  stated  (§  70)  that  to  calculate  the  sine  of 
any  angle  given  in  radian  measure  and  selected  at  random, 
one  must  use  the  relation  : 

sin0  =  0-^  +  ^-etc. 

[3      [5 

There  are,  however,  a  few  angles  whose  sines  can  be  deter- 
mined quite  readily  without  the  use  of  this  formula. 

Prominent  among  these  are  the  .ingles  whose  terminals 
pass  through  the  vertices  of  the  regular  polygons  of  3,  4,  6, 


120 


PLANE   TRIGONOMETRY. 


[§75 


8,  12,  sides,  the  polygons  being  central  at  the  origin,  with 
one  vertex  on  the  initial  line.  Such  terminals  will  form  the 
bisectors  and  trisectors  of  the  quadrantal  angles,  together 


with  the  border  lines  of  the  quadrants,  in  all  sixteen  termi- 
nals, as  in  the  above  diagram,  in  which  the  terminals  are 
indicated  by  numerals  and  also  in  degree  measure : 

(a)  For  terminals  1,  5,  9,  13,  the  sines  have  already  been 

given  (§  69).  The  student 
may  restate  the  results,  giving 
the  general  formulas.  Try  it 
without  looking  at  the  refer- 
ence, and  then  compare  results 
with  the  reference. 

(5)  For  terminals  3,  7,  11, 
15,  use  the  diagram  adjoining, 
in  which  a  right-angled  tri- 
angle whose  hypothenuse,  co- 
incident with  terminal  3,  is 
2  units  of  length,  and  taken 
with  one  vertex  at  the  origin 
and  one  leg  (length  V2)  on 
the  initial  line. 


§75c] 

(i)  Then,      sin  45°  = 


THE   SINE   FAMILY. 
MP 


121 


OP 


V2  .      7T 

2-  =  smT 


(ii)  Terminals  3,  7,  are  symmetric  to  the  vertical,  and 
therefore  belong  to  angles  of  the  same  sine  (§  62), 

.-.  sin  135°  =  sin  45°  =  ±^?  =  sin  ?  =  sin  — . 

2  4  4 

(iii)  Terminals  3,  11,  are  opposite  and  therefore  belong 
to  angles  of  opposite  sine  (§  63), 

.-.  sin  225°  =  -  sin  45°  =  -^  =  _  sin  Z  =  sin  — • 

2  4  4 

(iv)  Terminals  3,  15,  are  symmetric  to  the  horizontal,  and 
therefore  belong  to  angles  of  opposite  sine  (§  63), 


.-.  sin  315°  =  -  sin  45°  = 


V2 


.      IT  7  7T 

sin  —  =  sin— — 
4  4 


EXERCISES. 

1.  Find  the  sines  of  the  negative  angles  numerically  equal  to  those 
above. 

2.  Give  in  degree  measure  and  in  radian  measure  the  general  formulas 
for  all  angles  having  the  same  sine  as  45°.  Also  the  general  formulas  for 
all  angles  having  the  opposite  sine  of  45°. 

3.  Find  from  the  diagram  (Fig.  40)  six  positive  angles  having  the 
same  sine  as  45°.  Which  lines  may  be  their  terminals?  What  values 
of  m  in  the  general  formula  give  the 

angles  which  you  have  found?  150°  30° 

4.  Determine  similarly  six  negative 
angles. 

0)  For  terminals  2,  8,  10,  16, 
use  the  diagram  adjoining,  in 
which  an  equilateral  triangle  whose 
side  is  of  length  2,  is  set  with  one 
vertex  at  the  origin  and  a  bisector 
on  the  initial  line.  The  side  OP 
(Fig.  41)  will  fall  on  terminal  2, 
MP  will  be  1,  and  OM  will  be  VS.  FlG.  41. 


210 


122  PLANE   TRIGONOMETRY.  [§  75  c 

Then,  exactly  as  in  the  preceding  case, 
(i)     sin    30°=l  =  sin|; 

(ii)    sin  150°  =  sin  30°  =  J  =  sin  -  =  sin^ ; 

(iii)  sin  210°  =  -  sin  30°  =  -  £  =  -  sin  %  =  sin- 
v    j  2  6  6 


(iv)  sin  330°  =  -  sin  30°  =  -  £  =  -  sin  ^  =  sin 


7tt 
6 

11  7T 


EXERCISES. 
Take  those  on  page  121,  replacing  45°  by  30c 


(c?)  For  terminals  4,  6,  12,  14,  use  Fig.  42,  an  equilateral 
triangle,  whose  side  is  of  length  2,  one  side  coinciding  with 
the  initial  line  and  another  with  terminal  4. 

/-•\  •  r»AO  +v3  .     7T 

(l)    sin    60°  =  ^—  =  sin-; 


(ii)    sin  120°  =  sin  60°  = 


+  V3 


.        7T  •       2  7T 

=  sin  —  =  sin  -— — ; 
3  3 


(iii)  sin  240°  =  -  sin  60 


-V3 


2 

•       7T  .       4  7T 

-sin-=sin_; 
-V3 


(iv)  sin  300°  =  -  sin  60' 


.       IT  .      5  7T 

=  sm-=sm_. 


EXERCISES. 

Those  on  page  121,  replacing  45°  by  60°. 

Note.  —  The  student  is  advised  against  attempting  to  memorize  the 
foregoing  results  as  mere  facts.  Let  the  diagrams  be  called  up  when 
needed,  and  read  from  the  mental  diagram  the  value  sought : 


§75] 


THE   SINE   FAMILY. 


123 


(a)  For  45°  and  the  related  angles,  use  the  diagram  shown  in  Fig.  43. 

(b)  For  30°  and  the  related  angles,  use  the  diagram  shown  in  Fig.  44. 

(c)  For  60°  and  the  re- 
lated angles,  use  the  dia- 
gram shown  in  Fig.  45. 

These  diagrams  deter- 
mine the  sine  in  magni- 
tude. The  sign  is  then 
determined  by  the  loca- 
tion of  the  terminal. 


Fig.  44. 


Fig.  45. 


These  results  may  be  tabulated  as  follows : 

f  degrees  ]  0, 30,    45,        60,     90,   120,    135, 150,  180, 


I  radians  j  0,  -,      -,  -,       -,     — ,      "jr.  -J-,'* 

,  n.\    +V2    +V3    -     +V3    +V2   ,       A 

whose  sines  are   0,  J,  ,  — — ,  1,  — — ,— — ,  J,     0. 


EXERCISES. 

1.  What  are  the  sines  of  the  corresponding  negative  angles? 

2.  Read  from  mental  diagrams  the  sines  of  0°,  45°,  30°,  60°,  90°. 

3.  Considering  the  terminals  of  the  angles  in  Ex.  2  as  the  funda- 
mental terminals,  state  the  relative  position  of  any  other  one  of  the 
sixteen  terminals  of  Fig.  39  with  reference  to  some  one  of  these,  and 
thereby  determine  at  once  the  sine  of  the  corresponding  angle. 

The  class  may  be  drilled  several  days  in  succession  at  this,  until  the 
sine  of  any  one  of  the  sixteen  terminals  can  be  read  off  at  once  from  the 
proper  mental  diagram. 

4.  Using  the  principal  angle  corresponding  to  each  of  the  following 
sines,  determine  the  general  angles  for  the  following  expressions : 


sin 


sin- 


2 


un-M  --J+shr1 


(-«. 


3  sin"1 


5  sin"1!; 


(-•#♦«**$ 
(-*. 


~x  — -  +  2  sin 


«-& 


^..(-^^(-a 


124  PLANE   TRIGONOMETRY.  [§75 

sin-i(-i)  +  sin(-^?);  sin-i(-l)  +  3sin-i(-i)-2sin-i^; 

sin-1 0  +  sin-1! ;  2  sin"1 0 ; 

3sin-1l  +  2sin-1(l)-4sin-1(-^);6sin-1(^)-5sin-1(^^); 

and  so  on,  until  the  sines  of  these  fundamental  angles  are  known. 

5.  Verify  the  following  statements  (sign  not  considered)  : 

sin   2.45°  <  2  sin  45°;  sin   2.30°  <  2  sin  30° ; 

sin  2.90°  <  2  sin  90° ;  sin  2.60°  <  2  sin  60° ; 

sin  2.120°  <  2  sin  120° ;  sin  2.135°  <  2  sin  135° ; 

sin  2.150°  <  2  sin  150° ;  sin  2.180°  =  2  sin  180°. 

6.  Consider  similarly  the  remaining  positive  angles  less  than  360° 
corresponding  to  the  sixteen  terminals  of  Fig.  39. 

7.  Compare  results  when  the  angles  of  Ex.  5  are  taken  as  negative. 

8.  Give  three  special  solutions  and  the  general  solution  of : 

sin2  a;  =  0 ;  sin2  a:  =  1 ;  sin2  a;  =  \ ;  sin2a;  =  f ; 

sin2x  =  i;  sin3a;  =  i;  sin2—  =  ?;        V2sin5a?=l. 

4'  2  2       4 

9.  Using  the  general  formula  for  angles  of  a  given  sine,  determine 
the  general  values  of  the  angles  represented  by  the  expressions  in  Ex.  4. 

10.   Taking  the  general  form  of  the  quadratic  equation  as  ax2  +  bx 
+  c  =  0,  show  that  the  general  values  of  x  are 


_  _  b  ±  Vb2  -  4  ac 
2  a 
Sufficient  special  numerical  examples  may  here  be  assigned  to  make 
sure  that  the  class  can  write  at  once  the  solution  of  any  quadratic  equa- 
tion without  going  through  the  usual  process  of  completing  the  square. 

11.   Find  general  expressions  for  the  angles  satisfying  the  following 
equations : 

4  sin2  x  -  2  (1  +  V2)  sin  x  +  V2  =  0. 

4  sin2  x  +  2  (1  +  V2)  sin  x  +  V2  =  0. 
4  sin2a;  +  2  (1  -  V2)  sin  x  -  V2  =  0. 
4  sin2 x  +  2  (V2  -  1)  sin  x  -  V2  =  0. 
4  sin2 a;  -  2  (V3  +  1)  sin  x  +  V3  =  0. 
4sin2a:  +  2(\/3  +  1)  sin*  +  V3  =  0. 
4  sin2  a:  +  2  (V3  -  1)  sin  x  -  V3  =  0. 


§76]  THE  SINE  FAMILY.  125 

4  sin2  a;  -  2  (  V3  -  1)  sin  x  -  V3  =  0. 
4  sin2 a;  -  2  (\/2  +  V3)  sin  x  +  V6  =  0. 
4 sin2x  +  2  ( V2  +  a/3)  sin x  +  V6  =  0. 
2  V2  sin2 a:  -  V2  (V2  -  V3)  sinx  -  V3  =  0. 
2  V2  sin2 a;  -  V2  (V3  -  V2)  sina;  -  V3  =  0. 
2  sin2  x  -  ( V2  +  2)  sin  a:  +  V2  =  0. 
2sin2x+(\/2  +  2)sina;+\/2  =  0. 
2sin2a:+(>/2  -  2)  sin  a:  -  V2  =  0. 
2  sin2 a:  -  (V2  -  2)  sin  x  -  V2  =  0. 

12.  Find  what  values  of  x  will  satisfy  the  following  equations,  giving 
x  in  degrees  and  in  terms  of  it  in  each  case : 

sin"1  (a;2  -  x)  =  30° ;  sin"1  (x2  -  x)  =  ±  180° 

sin-1  (a:2  -  5  x)  =  -  30° ;  sin"1  (a;2  -  x)  =  ±  210° 

sin"^-  4a:2  +  7x)=  -  45°;  sin"1  (a;2  -  2ar)  =  ±  90° 

sin"1  (a;2  +  6  x)  =  60° ;  sin"1  (4  x2  -  x)  =  ±  270° ; 

sin"1  (3  x2  -  8  x)  =  -  60° ;  sin"1  (7  x2  -  x)  =  ±  225° ; 

sin"1  (4  x2  -  4  x)  =  120° ;  sin"1  (2  x2  -  3  x)  =  ±  240° ; 

sin"1  (3  x2  +  7  x)  =  -  120° ;  sin"1  (3  x3  +  5  x)  =  ±  360°; 

sin"1  (2  x2  -  4  x)  =  ±  135° ;  sin-1  (x2  -  x)  =  ±  300°. 
sin_1(x2  +  x)  =  ±  150°; 

13.  Name  two  angles  for  each  of  the  examples  of  Ex.  12,  which 
might  replace  the  second  member  of  the  equation  without  altering 
the  final  result. 

14.  Show  that,  knowing  the  sines  of  30°,  45°,  60°,  one  can,  by  con- 
structing a  diagram,  find  the  sines  of  the  following  angles: 

±405°;  ±390°;  ±420°;  ±840°;  ±660°;  ±690°;  ±1050°;  ±960°; 
±1305°;  ±1200°;  ±1020°;  ±945°;  ±855°;  ±870°;  ±810°;  ±900°; 
±780°;  ±570°. 

§  76.     Calculations  using  Sines,  without  use  of  Tables 
except  as  Check. 

Since  the  three  quantities,  sine,  ordinate,  modulus,  are 
connected  by  the  relation  of  definition, 

ordinate 

sine  = — T~r~ ' 
modulus 

if  any  two  are  given,  the  third  can  be  calculated  as  follows : 


126  PLANE   TRIGONOMETRY.  [§76 

(i)  When  ordinate  and  modulus  are  given, 

_  ordinate 
modulus 

(ii)  When  angle  and  modulus  are  given, 

ordinate  =  modulus  times  sine. 

(iii)  When  angle  and  ordinate  are  given, 

ordinate 


modulus  = 


sine 


These  three  formulas  are  to  be  memorized  each  as  a  primary  relation, 
so  there  will  be  no  hesitation  in  writing  any  one  of  them,  due  to  halting 
to  deduce  it  from  some  other.  The  author's  experience  is  that  where 
students  learn  trigonometry  from  the  ratio  definitions,  the  tendency  is  to 
hold  in  mind  as  a  primary  relation  that  of  (i),  there  being  a  halt  always 
in  seeing  (ii),  (iii),  due  to  their  deduction  from  (i).  The  class  should 
be  drilled  on  exercises  like  the  following  until  the  three  relations  are 
known  equally  well. 

EXERCISES. 

N.B.  Never  carry  the  number  of  figures  in  calculated  results  beyond  the 
number  in  the  given  quantities.  All  measured  sides  should  show  the  same 
number  of  significant  figures.  All  lengths  and  angles  given  must  be  assumed 
to  represent  measurements  and  are  thus  approximate. 

1.  Draw  diagrams  to  scale,  illustrating  the  following  data,  and  calcu- 
late the  corresponding  ordinates.  Test  each  result  by  calculating  back- 
ward the  given  modulus  from  the  calculated  ordinate.  Also  draw  the 
ordinate  in  the  diagram,  measure  it,  and  compare  the  result  with  the 
calculated  value.  Measure  the  angles  in  the  diagram  and  check  by  com- 
paring the  given  (or  calculated)  sine  with  the  table-sine. 

Modulus.       Sin  A.  Ordinate  ?  Modulus.  A°.  Oedinate  ? 

250  ±f  316  ±30 

350  ±|  250  ±45 

32  ±\  216  ±60 

16  ±\  215  ±90 
85  ±\  16  ±120 
90  ±\  80  ±135 

100  ±|  100  ±150 

25  ±J  120  ±210 

17  ±|  150  ±225 


76]  THE   SINE   FAMILY.  127 


Modulus. 

Six  A.           Ordinate?               Modulus. 

A°.          Ordinate  ? 

50 

±i 

30 

±240 

8 

±1 

70 

±300 

10 

±1 

80 

±315 

25 

±1 

90 

±330 

80 

±f 

100 

±360 

20 

±1 

200 

±390 

60 

±* 

8 

±405 

75 

±1 

10 

±420 

80 

dbA 

12 

±450 

100 

±A 

14 

±270 

200 

±1 

18 

±0 

2.  Interchange  the  words  "  ordinate  "  and  "  modulus  "  in  the  above 
table,  construct  to  scale  the  corresponding  diagram,  calculate  the  corre- 
sponding moduli,  test  as  in  Ex.  1 ;  measure  ;  compare. 

Note.  —  Construction  to  scale  will  be  found  quite  helpful  in  fixing  in 
mind  the  sine  as  a  ratio.  The  actual  lines  of  the  diagram  will  be  seen  to 
bear  to  each  other  the  ratio  indicated  by  the  sine. 

3.  Construct  to  scale  diagrams  illustrating  the  following  data,  and 
calculate  the  corresponding  sines  to  as  many  decimal  places  as  the  data. 

Sin  A.  Modulus.  Ordinate.  Sin  A . 


ODULUS. 

Ordinate. 

5 

±    3 

8 

±    4 

13 

±12 

13 

±    8.2 

17 

±15 

17 

±17 

17 

±    7.1 

25 

±24 

29 

±20 

2.9 

±    2.1 

37 

±12 

37 

±35 

41 

±    9.2 

41 

±19 

53 

±28 

5.3 

±    4.5 

Modulus. 

Ordinate. 

61.2 

±60.1 

61 

±11 

6.53 

±    3.32 

65 

±56 

65,24 

±  16.32 

6.5 

±    6.3 

73 

±48 

7.31 

±48.2 

0.85 

±    0.36 

85 

±77 

36 

±    9.2 

85.3 

±84.2 

89 

±80 

9.721 

±    6.532 

61 

±16 

43 

±34 

4.  On  the  diagrams  representing  the  data  above,  measure  any  other 
ordinate  and  its  modulus;  divide,  and  compare  the  resulting  sine  with 
that  gotten  from  the  given  numbers.  Also  measure  the  angle  and  com- 
pare calculated  sine  with  the  table  sine. 


128  PLANE   TRIGONOMETRY.  [§77 

§  77.     Selecting  and  Cutting  a  Table  of  Sines. 

As  already  indicated,  the  number  of  angles  for  which  the 
sine  can  be  determined  directly  from  a  diagram  is  quite 
limited,  and,  in  fact,  such  angles  here,  as  in  other  text-books, 
are  given  far  more  prominence  than  they  deserve.  It  is 
customary  to  use  the  tables  for  all  angles  except  multiples 
of  90°.  What  place  table  to  use  depends  upon  the  accuracy 
with  which  the  measurements  have  been  made  and  upon 
the  accuracy  to  which  the  calculated  result  is  desired.  As 
pointed  out  in  §§  20,  22,  24,  the  tendency  of  calculation 
manipulations  is  to  make  the  calculated  results  less  accu- 
rate than  the  measurements  on  which  they  are  based.  The 
resulting  inaccuracy  depends  on  the  number  of  additions, 
multiplications,  etc.,  made. 

It  must  be  assumed  in  calculations  with  triangles  that  the 
lines  whose  lengths  are  given  have  been  actually  measured, 
as  also  the  angles  given.  All  the  given  parts  of  a  diagram 
should  indicate  about  the  same  degree  of  accuracy  in  meas- 
urement. 

In  calculating  triangles,  the  number  of  operations  being 
always  few,  the  following  rules  may  be  laid  down : 

(1)  All  measured  lines  must  show  the  same  number  of 
significant  figures.     (See  §  33.) 

(2)  Calculated  lines,  sines,  and  logarithmic  sines  must 
not  show  more  significant  figures  than  the  measured  lines. 
They  may  generally  show  the  same  number.     (See  §  30.) 

(3)  When  the  measured  lines  of  a  diagram  show  only  one 
significant  figure,  measured  angles  and  calculated  angles  may 
read  to  the  nearest  five  degrees;  with  only  two  significant 
figures  in  lines,  angles  may  read  to  the  nearest  half-degree, 
as  with  a  surveyor's  compass. 

(4)  When  the  measured  lines  of  a  diagram  show  only 
three  significant  figures,  angles  may  read  to  the  nearest  five 
minutes ;  while  with  only  four  significant  figures,  the  meas- 
ured angles  and  the  calculated  angles  may  read  to  the  near- 


§79]  THE   SINE   FAMILY.  129 

est  minute  when  a  four-place  table  is  used,  and  to  the  nearest 
ten  seconds  when  a  five-place  table  is  used. 

(5)  In  cases  (3),  (4),  a  four-place  table  is  sufficiently 
accurate,  generally.  At  times  four  significant  figures  in 
measured  results  call  for  the  use  of  five-place  tables.  (See 
§  23.)  Whatever  table  is  used  should  be  cut  back  to  one 
place  more  than  the  data. 

(6)  When  the  measured  lines  show  five  significant  figures, 
a  five-place  table  or  a  six -place  table  should  be  used.  The 
measured  angles  and  calculated  angles  may  then  read  to  the 
nearest  second  with  a  six-place  table,  and  to  the  nearest  five 
seconds  with  a  five-place  table. 

(7)  In  astronomical  calculations,  when  the  number  of 
terms  makes  a  balancing  of  table-errors  likely,  a  five-place 
table  is  used  when  the  given  angles  and  calculated  angles 
are  given  to  the  nearest  second ;  a  six-place  table,  when 
tenths  of  seconds  appear;  a  seven-place  table,  when  hun- 
dredths of  seconds  appear. 

(8)  Do  not  use  a  seven-place  table  when  a  four-place  table 
is  sufficient,  nor  a  four-place  table  when  a  seven-place  table 
is  necessary.  One  place  beyond  the  data  is  all  that  is  ever 
needed.  The  smaller  the  suitable  table,  the  greater  is  the 
saving  of  time  in  manipulation. 

Proof  of  the  Rules. 

Sin  #  =  0.1  means  that  sin  #  lies  between  0.05  and  0.15. 
Taking  from  the  tables  the  angles  opposite  the  sines  0.05 
and  0.15,  which  stand  closest  to  the  sine  0.1,  we  get  3°  26' 
and  8°  38',  whose  mean  may  be  taken  as  x  when  sin#=  0.1, 
and  half  whose  difference,  or  2°  36',  will  be  the  possible 
error  in  x.  Similarly,  sin  x  =  0.2  has  as  extremity  angles, 
9°  15'  and  14°  29',  whose  mean  may  be  taken  as  #,  with  the 
possible  error  of  2°  37'.  Treating  0.3,  0.4,  •••,  0.9,  the  same 
way,  the  average  error  for  the  whole  table  is  about  3°  30'  in 
determining  an  angle  from  one  figure  in  the  sine.  Treating 
0.10,  0.20,  etc.,  in  the  same  way,  the  average  is  about  20'. 


130  PLANE   TRIGONOMETRY.  [§78 

Safe  working  limits  then  are  5°  and  30',  respectively,  as  in 
rule  (3).  An  increase  of  one  figure  in  the  line-data,  when 
the  table  has  at  least  one  place  more  than  the  data,  divides 
the  error  by  10.     This  and  §  23  give  the  remaining  rules. 

§  78.     A  Four-place  Table. 

In  the  tables  accompanying  this  book  is  a  four-place  table  of  sines 
and  of  logarithmic  sines.     It  is  not  to  be  used : 

(1)  When  the  measured  lines  of  the  diagram  show  more  than  four 
significant  figures. 

(2)  When  the  measured  angles  of  a  diagram  show  seconds. 

(3)  For  small  angles  (less  than  about  5°)  when  the  sines  (angles)  can- 
not be  obtained  directly  from  the  table  without  interpolation.  Here  the 
sines  change  too  rapidly  to  use  their  differences.     (See  §  186.) 

How  to  use  these  tables  is  explained  in  the  preface  of  the  tables. 

In  practical  work  in  electrical  and  mechanical  engineering  the  data 
are  largely  three-figured.  A  four-place  table  meets  most  of  the  re- 
quirements. 

§  79.     Some  Calculations  using  Sines  of  the  Tables  and 
Logarithmic  Sines.     (Four-place  Tables.) 

From  the  relations 


(i) 

Sme           = JT-' 

modulus 

(ii) 

ordinate  =  modulus  times  sine, 

(iii) 

-,   i          ordinate 
modulus  = : , 

sine 
follow  the  relations 

(iv)  log  sin    =  log  ord  —  log  mod, 

(v)  log  ord   =  log  mod  +  log  sin, 

(vi)  log  mod  «s  log  ord  —  log  sin. 

A.  Examples  on  Calculating  the  Ordinate  (case  ii). 

(i) 

-,.        I  modulus  =    8         ^^  *lnc\. 

GlV6n  I  angle       =  25°  -  ordlnate 

Fig.  46. 


§79]  THE   SINE   FAMILY.  131 

Ordinate  =  modulus  times  sine, 
log  ord     =  log  mod   +  log  sin. 

Modulus  being  given  to  only  one  significant  figure,  the 
ordinate  should  be  calculated  to  only  one  significant  figure. 
By  natural  sines,  to  as  many  figures  as  the  data,  is  the  proper 
method.  We  give  logs  here  and  in  (2)  as  a  mere  check,  or 
rather  as  a  mere  sample  process  in  cutting  logs  to  one  figure 
beyond  the  data. 

SOLUTION   BY   NATURAL   SINES.  SOLUTION   BY   LOGARITHMS. 

sin  25°  =  0.4  log  8           =  0.90 

mod      =  8_  log  sin  25°  =  1.63 

.-.  ord        =  3  log  ord        =  0.53 

Possible  error  =  ±  0.6  .*.  ord        =  3 


(2)  Given,  modulus  =  27,  angle  =  22°. 

HON  BY  NATURAL  SINES.  SOLUTIC 

a  22°=    0.38  1 

mod  =  27 log  sin  22°  =  1.574 

7.6  log  ord  =  1.005 

2.6  ...  ord  =10 


SOLUTION  BY  NATURAL  SINES.  SOLUTION  BY  LOGARITHMS. 

sin  22°=    0.38  log  27  =  1.431 


10         ±  0.3 

The  solution  by  natural  sines  is  to  be  carried  out  by  the 
shortened  process  of  multiplication.  (See  §  26.)  The  num- 
bers 0.6  and  0.3  after  the  ordinates  show  the  possible  error  in 
these  results  if  tenths  were  written  down,  and  are  obtained 
by  §  25. 

The  student  is  advised  to  follow  some  scheme  like  that 
above.  Place  the  data  on  one  side  of  a  diagram.  Place  the 
things  to  be  found  on  the  other  side.  Draw  the  given  parts 
of  the  diagram  in  heavy  lines  ;  the  parts  to  be  found  in  dots. 
Make  the  diagram  to  scale  on  coordinate  paper  ;  measure  as 
a  check.  Place  the  formulas  to  be  used  under  the  diagram. 
Check  by  combining  the  given  and  calculated  quantities  in 
some  new  relation,  directly  or  by  logarithms. 


132  PLANE   TRIGONOMETRY.  [§79 

Note.  —  The  student  may  solve  the  following  examples  both  by  natu- 
ral sines  and  by  logarithmic  sines,  and  compare  results.  He  may  also 
work  the  examples  backward,  as  a  test  of  accuracy.  Since  every  calcula- 
tion ought  to  be  tested  in  some  way  by  the  person  making  it,  no  answers 
are  given.  With  one-figured  and  two-figured  data,  practical  engineers 
use  the  natural  functions. 

EXERCISES. 

Make  all  diagrams  to  scale  on  coordinate  paper.  Measure  the  ungiven 
part  as  a  check  on  calculation. 

Keep  in  mind,  in  making  the  diagrams,  that  with  data  like  Ex.  1,  a 
measured  modulus  9  means  not  9.0,  but  something  between  8.5  and  9.5, 
and  that  the  angle  35°,  given  as  going  with  such  a  modulus,  means  that 
35°  is  the  angle  to  the  nearest  5°,  and  that  the  angle  lies  between  30°  and 
40°,  but  nearer  35c  than  to  30°  or  to  40°.  If  diagrams  are  made  repre- 
senting the  extreme  cases  of  measurement  signified  by  the  data,  it  will 
become  very  apparent  why  calculated  lines  should  not  be  carried  to  more 
places  than  the  data.  Make  such  diagrams  for  the  following  examples. 
In  examples  like  5,  6,  7,  8,  modify  the  angle  readings  to  suit  the  table 
used,  according  to  the  rules  on  page  129. 

1.  Modulus  9,  angle  ±  35° ;  modulus  7,  angle  ±  55° ;  modulus  6, 
angle      85°. 

2.  Modulus  9,  angle  ±  105° ;  modulus  8,  angle  ±  200° ;  modulus  7, 
angle  ±  305°. 

3.  Modulus  23,  angle  ±  34° ;  modulus  5.6,  angle  ±  123° ;  modulus  6.8, 
angle  ±  196°  30' ;  modulus  84,  angle  ±  302°  30' ;  modulus  23,  angle 
±4325°. 

4.  Modulus  45.3,  angle  ±  34°  5' ;  modulus  4.57,  angle  ±  174°  10' ; 
modulus  34.8,  angle  ±  267°  20' ;   modulus  41.3,  angle  ±  27942°  15'. 

5.  Modulus  43.41,  angle  ±  23°  14' ;  modulus  4.587,  angle  ±  174°  23' ; 
modulus  0.4752,  angle  234°  43' ;   modulus  678.4,  angle  ±  4545°  2'. 

6.  Modulus  345.61,  angle  34°  25'  43" ;  modulus  23.598,  angle  172° 
34'  56" ;  modulus  4567.8,  angle  256°  31'  36" ;  modulus  0.58923,  angle 
295°  17'  13" ;  modulus  7.8519,  angle  862°  44'  44"  (at  least  five-place  table). 

7.  Modulus  234.567,  angle  45°  47'  53".2  (at  least  six-place  table). 

8.  Modulus  2564.321,  angle  82°  13'  27".34  (at  least  seven-place  table). 

9.  What  data  would  suit  an  uncut  ten-place  table  ? 
10.   What  data  would  suit  a  three-place  table  ? 


79] 


THE   SINE   FAMILY. 


133 


B.    Examples  on  Calculating  the  Modulus  (case  iii). 

(i) 

Given 


( ordinate  =  8 


1  angle 


17°   ^^yn* 

A  Fig.  47. 


Modulus 


0 

ordinate 
sine 


Find 

modulus  =  30 


.*.  log  mod  =  log  ord  —  log  sin. 

SOLUTION   BY   SINES.  SOLUTION  BY  LOGARITHMS. 


sin 

.3)8 
26 
which  must  be  taken  as   30, 
since  ordinate  has  only  one 
significant  figure.     Are  the 
data  consistent  ? 

(2) 
ns„„„  (ordinate  =  13. 27 
GlVen  (angle      =42°17' 

A 


log  8  =  0.90 

log  sin  17°  =  1.47 

log  mod  =  1.43 

.-.  mod  =  27 
.-.  mod  =  30 


A 


4>y 


v42  17 


g;        Find 

9     modulus  =  19.73 


Modulus 


Fig.  48. 
ordinate 


sine 
log  mod  =  log  ord 


log  sin. 


SOLUTION  BY  NATURAL  SINES. 

(Shortened  division  process.) 

sin  19.73  mod 

6728)132700 

673 

654 

605 

49 

47 

2 


SOLUTION  BY  LOGARITHMS. 

log  ord  =1.1229 
log  sin  =  1.8279 
log  mod  =  1.2950 

mod  =  19.73 


134  PLANE   TRIGONOMETRY.  [§79 

The  solution  by  natural  sines  is  made  by  the  shortened 

process  of  division.    (See  §  40.)    To  determine  to  how  many 

places  the  division  may  be  carried,  use  may  be  made  of  §  27 

and  §  28,  from  which  it  follows  that  the  per  cent  rejection 

error  in  the  modulus  is  1§^-  (g yV ¥  +  TFT2")  Per  cen^?  or  about 

KeV  +  iV)>  or  about  KeV  +  tV)>  or  about  A  of  1  Per  cent' 
or  about  -fa  of  1  per  cent.     The  modulus  being  about  20,  the 

error  is  about  0.01.     The  division  should  not  be  carried, 

therefore,  beyond  hundredths.     By  the  shortened  process  of 

division,  the  operation  is  self-limited,  when  the  data  have 

the  same  number  of  significant  figures.    It  ceases  at  the  limit 

of  possible  inaccuracy. 

The   results   by  natural   sines   and   by  logs   might  have 

differed  by  1  in  the  last  place. 

The  work  may  also  be  tested  by  the  backward  calculation, 

ordinate  =  sine  times  modulus. 

0.6728  sin 

19.73       mod 

6.728 

'     6.055 

.470 

.020 


13.27       ord 

EXERCISES. 

Interchange  the  words  "ordinate"  and  "modulus"  in  the  exercises  of 
(A),  page  132,  and  solve.  Determine  in  each  case  the  maximum  error  in 
the  result  due  to  rejection  error  in  the  ordinate  and  sine.  Make  dia- 
grams to  scale  on  coordinate  paper,  and  measure  the  ungiven  parts  as  a 
check  on  calculation. 

C.     Examples  on  Calculating  the  Sine  and  Angle 
from  the  Ordinate  and  Modulus  (case  i). 

f  ordinate  =  36.16  4$^         S  Find 

1VGn  I  modulus =67.32  S^ t  *      angle  =  31°  29' 

O  M 

Fig.  49. 


79] 


THE  SINE   FAMILY. 


135 


_  ordinate 
modulus 
log  sin  =  log  ord  —  log  mod. 

SOLUTION   BY   LOGARITHMS. 

log  ord  =  1.5582 
log  mod  =  1.8281 


log  sin  =  1.7301 

Table  angle,  32°  29'. 

General  angle, 

m.180  +  (-rr32°29'. 


SOLUTION   BY   NATURAL    SINES. 
(Shortened  division.) 

0.5371  sin 
6732)3616. 
3366 
•      250 
202 
48 
47 
1 
Table  angle,  32°  29'. 
General  angle, 
m.  180° +  (-1)™ 32°  29'. 

ADDITIONAL   TEST   OF    ACCURACY. 

ordinate  =  modulus  times  sine. 
67.32      mod 

.5371  sin 
33.660 
2.020 
.471 
.007 
36.16    =ord 
To  determine  the  number  of  places  to  which  the  sine  may- 
be determined  from  the  data,  assuming  rejection  error,  use 
may  be  made  of  §  27  and  §  28. 
Per  cent  rejection  error  in  sine  is 

4*  CnW  +  t&f>  or  about  i  (eV  +  A)i 

or  about  \  (y1^  +  ^),  or  about  if  ^  of  1  cjo. 

Now,  ^  of  1%  of  the  sine,  0.5371,  is  about  0.0001.  So 
the  division  may  not  be  carried  beyond  four  figures. 

The  results  here  and  in  the  previous  cases  agree  with  the 
general  statement  made  in  rule  (2)  of  §  77,  that  calculated 


136  PLANE   TRIGONOMETRY.  [§79 

results  must  not  be  carried  beyond  the  number  of  significant 
figures  in  the  data. 

It  will  be  observed  in  any  large  table  of  sines,  where 
seconds  are  taken  into  account,  that  over  a  large  part  of  the 
table,  when  the  angles  differ  by  only  a  second,  the  sines 
agree  in  the  first  four  figures.  Thus,  in  four-figured  data, 
calculated  angles  cannot,  as  a  rule,  be  determined  to  the 
nearest  second.     It  is  sufficient  to  give  only  minutes. 

EXERCISES. 

Determine  the  sine  and  angle  in  the  following  cases : 

Ordinate.         Modulus.  Ordinate.         Modulus.  Ordinate.        Modulus. 

72.16  95.73  71.6  94.3  2.31  8.56 

1.831  3.924  54.3  83.9  0.5673  1.906 

21  67  5  7  5.43  9.56 

Take  the  ordinates  as  both  plus  and  minus. 

§  80.   Using  a  Five-place  Table. 

Engineering  students  should  have  some  good  five-place  tables,  as  Gauss's 
or  Hussey's.  The  teacher  may  here  explain  the  use  of  such  a  table,  if  it 
is  to  be  used.  He  may  also  make  five-figured  data,  corresponding  to  the 
cases  above,  and  have  the  class  make  the  corresponding  calculations. 
Given  angles  will  read  now  to  seconds ;  also  calculated  angles  will  be  given 
to  seconds,  or  five  seconds.     See  §  77,  (6). 

Similarly,  some  practice  may  be  given  on  examples  where  a  six-place 
or  seven-place  table  is  called  for. 

§  81.    Solution  of  Right-angled  Triangles  by  Sines. 

The  selection  of  the  position  of  the  initial  line  is  altogether 
arbitrary,  as  is  also  the  direction  for 
positive  or  negative  turn.  If  in  Fig.  50 
AH  is  taken  as  initial  line  and  AB 
as  modulus,  with  the  counter-clockwise 
turn  as  positive,  then  HB  is  ordinate, 

.:smA  =  M. 
AB 

Similarly,  counting  BR  as  initial  line, 
BA  as  modulus,  and  the  clockwise  turn 
as  positive,  then  HA  is  ordinate, 


§81]  THE   SINE   FAMILY.  137 

••Bin  B  =  M. 
BA 

In  considering  right-angled  triangles,  it  is  not  customary  to 
pay  any  attention  to  the  question  of  direction,  either  of  turns  or 
of  lines,  but  to  use  the  general  statements 

_  side  opposite  the  angle 
hypothenuse 

The  angles  being  acute,  their  sines  are  always  positive. 
Using  A,  a-,  B,  b,  to  denote  angles  and  their  opposite  sides, 
or  pairs  of  opposites,  h  being  the  hypothenuse,  we  have : 

(1)  sinA  =  j;  sin  B=-,  or, 

h  h 

the  sine  of  either  angle  is  the  side  opposite  divided  by  the 
hypothenuse. 

(2)  a  =  smA-h;  b  =  sinB'h,or, 

either  side  is  the  sine  of  the  opposite  angle  times  the  hypothe- 
nuse. 

(3)  h  =  ^~7=:- — 5'  or' 

sin  A      sin  B 

the  hypothenuse  is  either  side  divided  by  the  sine  of  its  opposite 
angle. 

(4)  h=Vair+¥. 

The  relation  (4)  can  be  used  advantageously  to  determine 
h  only  when  a,  b  are  small.  After  h  is  found,  either  angle  can 
be  calculated  from  its  sine.  When  a,  b  are  large,  the  angles 
are  best  found  by  a  method  which  will  be  given  later. 
(See  §  169.) 

These  relations,  (1),  (2),  (3),  are  easily  memorized,  and 
should  be  memorized  thoroughly,  each  as  a  primary  fact. 
Note  that  in  (1),  (2),  opposites,  as  A,  a,  etc.,  are  on  opposite 
sides  of  the  relations.  Thus,  as  in  (1),  when  sin  A  is  written, 
write  at  once  on  the  other  side  of  the  equation  its  opposite, 
a,  and  then  under  a  write  h,  since  a  sine  is  a  ratio.  Like- 
wise, when  a  is  placed  on  one  side  of  the  equation,  write  the 


138  PLANE  TRIGONOMETRY.  [§81 

sine  of  its  opposite,  sin  A,  on  the  other,  and  then  h,  to  make 
the  relation  homogeneous,  —  a  line  equal  to  a  line. 

Examples  in  right-angled  triangles  can  be  of  only  four 
types,  corresponding  to  the  four  sets  of  relations  above : 

(a)  Given  the  hypothenuse  and  a  side,  to  find  the  two  angles 
and  the  remaining  side. 

For  example,  given  a,  h, 

then  sin^l  =  ^;  B=90-A;  b=smB-h. 

h 

(5)    Given  either  angle  and  the  hypothenuse,  to  find  the  other 

angle  and  the  two  sides. 

For  example,  given  A,  h, 

B  =  90  —  A  ;  a  =  sin  A  •  h  ;  b  =  sin  B  •  h. 

(tf)    Given  a  side  and  an  angle,  to  find  the  remaining  parts. 
For  example,  given  a,  A, 

h  =  -A-r;  B=90-A;  b^sinB'h. 
sin  A 

(c?)    Given  the  two  sides,  to  find  the  remaining  parts. 

h  =  ^aT+¥',  sin^.=  ^;   B  =  90-A. 
h 

See  note  above  in  connection  with  (4),  page  137. 

§  82.    Testing  the  Solution. 

With  A  +  B  =90°,  h  =  -JL-  =  -^—. 

sin  A     sin  B 

,\  a  •  sin  B  =  b  •  sin  A. 

h2-a2  =  P. 

Thus  the  log-checks : 

log  a  +  log  sin  B  =  log  b  -f  log  sin  A. 

log  (h-a)+  log  (h  +  a)  =  2  log  b. 

In  applying  log-checks  to  "answers,"  the  logarithms  of 
such  answers  must  be  looked  up  in  the  tables,  instead  of 
checking  with  the  logarithms  of  the  calculation  scheme.  The 
latter  may  be  correct  and  yet  the  "  answers  "  be  incorrect,  an 


§83]  THE   SINE   FAMILY.  139 

error  having  been  made  in  looking  up  the  numbers  corre- 
sponding to  logarithms. 

Logarithmic  checks  are  not  to  be  held  to  a  closer  agreement 
than  the  data  call  for.  It  is  always  sufficient,  as  pointed  out 
in  Chapter  II.,  if  they  agree  within  2  in  the  last  place,  such 
last  place  not  being  necessarily  the  last  place  of  the  table  used, 
but  the  place  to  which  the  results  are  called  upon  to  be  correct, 
in  accordance  with  the  data.  For  example,  if  the  data  are 
three-figured,  the  check  is  met  if  the  logs  differ  by  not  more 
than  2  in  the  third  place.     (See  the  example,  p.  140.) 

In  extended  astronomical  calculations,  where  the  number 
of  terms  is  sufficient  to  allow  for  balancing  of  errors,  it  is 
also  reckoned  sufficient  if  the  log-checks  agree  to  within  2 
in  the  last  place. 

EXERCISES. 

Test  the  following  triangles  for  consistency,  using  log-checks : 

1.  a  =  42,  b  =  56,  h  =  70 ;   a  =  231,  b  =  95.7,  h  =  250. 

2.  a  =  231,  A  =  67°  30',  b  =  95.7,  B  =  22°  30'. 

3.  a  =  229,  A  =  37°  20',  b  =  300,   B  =  52°  40'. 

§  83.     Solution  Scheme  for  solving  Right-angled  Triangles 
when  Logarithms  are  used. 

Construct  to  scale  a  diagram  on  which  the  parts  given  are 
indicated  by  heavy  lines,  and  the  parts  to  be  found,  by  dotted 
lines.  Write  on  the  given  parts  their  numerical  values.  Do 
the  same  with  the  dotted  parts,  after  they  have  been  found. 

Set  the  parts  given  in  one  column,  as  in  the  following 
example.  Set  off  a  similar  column  for  the  parts  to  be  found, 
in  the  order  in  which  they  will  be  found.  Fill  in  with  the 
parts  found  after  the  calculations  have  been  carried  out. 
Give  under  the  diagram  the  formulas  to  be  used. 

The  following  calculation  scheme  is  suitable  for  all  cases 
except  when  the  two  sides  are  given.  Merely  the  order  of 
filling  in  the  scheme  will  change  with  change  of  data,  the 
scheme  itself  remaining  unchanged. 


140 


PLANE   TRIGONOMETRY. 


[§83 


Model  Example. 


Given  !*  =  27-3 

I B  =  35°  15' 


Solution  formulas 


f  .4  =  54°  45' 
Find  J    ft  =  15.8 
I  a  =  22.3 


Fig.  51. 

A  =  90  -  B 

b  =  sin  B  •  A.  . 
a  =  sin  -4  •  h.  . 


LOGARITHMS. 

(4)  6  =  1.1975 

(2)  sinB  =  1.7613 
(1)  h  =  1.4362 

(3)  sin  A  =  1.9120 

(5)  a  =  1.3482 


log  b  =  log  sin  B  -f  log  ^ 
log  a  —  log  sin  J.  +  log  h 

Solution  Scheme. 

test. 
a  sin  B  =  b  sin  A 


LOGARITHMS. 


1.1987 
1.9120 


1.1107 


1.3483 
1.7613 


1.1096 


Here,  in  applying  the  checks,  we  have  taken  the  logs  of 
15.8  and  22.3,  the  nearest  three-figured  approximations  cor- 
responding to  the  logs  of  b,  a,  and  not  the  logs  of  5,  a  in  the 
scheme.  We  are  cheeking  our  answers.  The  data  being  three- 
figured,  the  answers  are  three-figured.  The  log-check  checks 
to  within  2  in  the  third  place,  which  is  all  that  is  required 
in  the  case  of  three-figured  data.     (See  §  82.) 

The  order  of  filling  in  the  solution  scheme  is  indicated  by 
the  numerals,  (1),  (2),  (3),  (4),  (5). 

First  look  up  logarithms  of  (1),  (2),  (3).  Then  add  (1), 
(2),  and  set  the  result  after  (4).  Next  add  (1),  (3),  and  set 
the  result  after  (5). »  Then  look  up  the  numbers  correspond- 
ing to  logarithms  (4),  (5),  to  as  many  places  as  the  data,  and 
enter  the  results  after  "  Find  "  in  the  calculation  scheme. 

It  is  generally  best  not  to  enter  numbers  on  the  solution 
scheme,  but  letters,  taking  their  logarithms  to  correspond  to 
their  values  in  the  data. 


§  84]  THE   SINE   FAMILY.  141 

If  the  given  parts  are  a,  A,  the  same  solution  scheme  is 
used,  but  the  order  of  rilling  in  would  be  as  indicated  by  the 
numerals  following : 

LOGARITHMS. 

(5)  b  = 

(3)  sin  B  = 

(4)  h  = 
(2)  sin  A  = 
(1) 

Enter  the  logs  (1),  (2),  (3).  Next  subtract  (2)  from  (1) 
for  (4).     Then  add  (3),  (4),  for  (5). 

EXERCISES. 

1.  Indicate  the  order  of  filling  in  for  any  other  case  which  may  arise. 

2.  Make  schemes  into  which  cologs  enter  where  subtractions  occur 
above.     Which  do  you  prefer  ? 

NUMERICAL  EXAMPLES. 

(Make  diagrams  to  scale,  for  check.) 

Determine  both  by  natural  sines  and  by  logarithms,  using  the  scheme 
given  above,  the  ungiven  parts  of  the  following  triangles,  testing  the 
final  results  as  indicated  above,  finding  also  to  how  many  significant 
figures  the  results  should  be  taken  if  the  data  are  assumed  subject  to 
rejection  error : 

1.  Given  £  =  234.5,  a  =  136.8;  £=234,  a  =  156;  £  =  23,  a  =  15. 

2.  Given  £  =  46.7,  6  =  23.8;  £  =  46,  a  =  25;  £  =  4,  a  =  2. 

3.  Given  £  =  9.68,  a  =  5.76;  £  =  95,  a  =  57;  £  =  9,  a  =  5. 

4.  Given  £  =  468.2,  ,4=46°  34';  £  =  465,  ,4  =  45°  35' ;  £  =  46,4=45°. 

5.  Given  £  =  34.43,  B  =  38°  23' ;  £  =  54.4,  4  =  56°  15' ;  £  =  5,  4  =  40°. 

6.  Given  £  =  543.2,  A  =68°  43';  £  =  543,  A  =66°  45';  £  =  54,4  =  80°. 

7.  Given  a =45.6,  4  =76°  35';  a  =  46,  4  =45°  30' ;  £=4,4=45°. 

8.  Given  6=213.4,  £  =  64°46';  6  =  21.3,  A  =54°  5' ;  £=21,  4=54°. 

9.  Given  a  =  96.2,  B  =  49°  45';  a  =  96,  5=49°  30';  £  =  9,5=40°. 

§  84.    Solution  with  Five-place  Table. 

The  teacher  may  construct  and  assign  corresponding  data  calling  for 
the  use  of  a  five-place  table. 


142 


PLANE   TRIGONOMETRY. 


[§85 


§  85.  Calculations  with  Sines,  with  Practice  Examples  in  the 
Use  of  Logarithms  and  the  Shortened  Processes  of  Multipli- 
cation and  Division  on  Exact  Data. 

Let  AHB  (Fig.  52)  be  a  right-angled 
triangle.  Drop  the  perpendicular  HO 
on  AB.  Then  BO  is  called  the  pro- 
jection of  a  on  AB,  and  will  be  denoted 
by  ar  Similarly  A  0  is  &'s  projection, 
or  bp.  The  angle  AHQ  is  the  same  as 
B,  while  BHQ  is  the  same  as  A. 


a„=  a  •  sin  CUB  =  a  •  sin  A 


-p 

h 

h 

h 

=  ft. 

sin 

OHA  = 

:b. 

sin 

B 

-I- 

b2 
=  h' 

h 

=  a 

•  sin 

CBH, 

or 

b- 

sin 

CAH  = 

ab 
~~  h 

CH 

(a)  Oral  exercises  in  connection  with 
the  preceding  triangle  : 

Taking  a,  5,  h  from  the  table  of 
triangles,  page  145,  as  in  the  adjoin- 
ing diagram  for  Ex.  1  of  that  table, 
excellent  practice  on  the  sine  may  be 
had,  as  follows  : 


a) 

(2) 
(3) 


a=3 


TEACHER  (QUESTION). 

sin^.? 
sin  B? 
sin  CHB  ? 
sin  CHA  ? 
CH? 

AC? 
BC? 


STUDENT  (ANSWER). 

I 

i 


=  sin  A  =  | 
=  sin  B  =  f 
=  sin  B  times  BH 
or,  sin  A  times  AH 


|  of  3  =  # 

fof4  =  -^ 


sin  OKI  times  AET  =  f  of  4  =  -^_ 


sin  CHB  times  ##  =  f  of  3 


The  exercise  may  be  varied  by  tilting  the  triangle  in 
various  attitudes,  until  the  student  is  thoroughly  familiar 
with  : 


§85  6] 


THE   SINE   FAMILY. 


143 


(i)  The  sine  is  the  side  opposite  divided  by  the  hypothe- 
nuse. 

(ii)  Any  side  is  the  sine  of  the  opposite  angle  times  the 
hypothenuse. 

(iii)  The  hypothenuse  is  any  side  divided  by  the  sine  of 
opposite  angle. 

(5)  Numerical  exercises  in  connection  with  the  triangle, 
Fig.  52,  and  the  table,  page  145  : 

In  addition  to  (1),  (2),  (3),  for  practice  in  the  use  of 
logarithms  and  in  the  shortened  processes  of  division  and 
multiplication,  one  may  calculate  the  area,  A,  of  the  triangle 
AHB,  and  the  radii  of  the  circles  tangential  to  the  sides  of 
this  triangle. 

The  area  of  AHB  is      AH '  BH 


or, 


ab      A 


(4) 


The  radius  of  any  tangential  circle  is  found  by  joining  its 
centre  to  the  vertices  A, 
H,  B  (Fig.  54).  In  the 
case  of  the  internal  circle 
the  sum  of  the  three  new 
triangles  formed  is  the 
original  triangle  AHB. 
In  the  case  of  an  external 
circle  (an  escribed  circle, 
it  is  generally  called),  the 
original  triangle  is  the 
sum  of  the  two  new  tri- 
angles which  have  for 
sides  the  two  sides  of  the 
original  triangle  touched 
on  the  border  of  the  orig- 
inal triangle,  minus  that  new  triangle  whose  side  is  that  side 
of  the  original  triangle  touched  on  the  border  of  the  domain 
outside  the  original  triangle. 


Fig.  54. 


144  PLANE   TRIGONOMETRY.  [§85  J 

For  the  internal  circle,  whose  centre  is  0, 

aCAH+aOHB  +  AOBA  m  AAHB, 
or,  r%(a  +  b  +  h)  =  a  •  5, 

1      a+b+h  K J 

For  the  escribed  circle  touching  AB  and  whose  centre  is  C, 
AC'AH  +  AC  KB  -  AC'BA  =  AAHB, 
or,  rA(a  +  5  —  Ji)  =  a  •  5, 

or,  rh  =  — — -.  (6) 

a  +  b  —  h 

EXERCISES. 

1.  Deduce  expressions  corresponding  to  (6)  for  ra,  rb. 

2.  Show  that  if  A  is  the  area  of  any  triangle,  right-angled  or  oblique, 
and  s  its  semiperimeter, 

A  A  A  A 

rb  = 


s  —  a  s  —  b  s  —  c  s 

where  ft  is  the  radius  of  the  inscribed  circle. 

3.  Show  that  -+-  +  —  =  -,  for  all  triangles. 

r«      rh      rc      rt 

4.  If  rQ  is  the  radius  of  the  circle  circumscribing  any  triangle, 

_       a  b       _       c       _  abc 

0  ~  2  sin  ,4  ~  2  sini>'  ~  2  sin  C  ~  4~A 

5.  Show  that  ry4  = — 

0        2(a  +  b  +  c) 

6.  Show  that  the  lengths  of  tangents  from  the  vertices  A,  B,  C  to 
inscribed  circle  are  s  —  a,  s  —  b,  s  —  c,  respectively ;  and  for  the  escribed 
circles  s,  s  —  b,  s  —  c ;  s,  s  —  c,  s  —  a ;  s,  s  —  a,  s  —  b. 

(c)  Table  of  triangles  (taken  from  Dr.  Conradt's  "  Trigo- 
nometry") for  use  in  connection  with  exercises  (#),  (6) 
preceding. 

The  table  on  page  145  gives  a  large  number  of  integers  for  which 
a2  +  b2  =  h2,  and  from  them  a  great  variety  of  exercises  on  sines  (and 
later  cosines  and  tangents)  will  suggest  themselves  to  the  teacher.  The 
numbers  here  being  exact,  and  not  representing  measurements,  calculated 
parts  can  be  carried  to  any  number  of  decimals. 


§85c] 


THE  SINE   FAMILY. 


145 


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146 


PLANE   TRIGONOMETRY. 


[§86 


Jh 


§  86.     Vertical  Projections. 

If  perpendiculars  AAV  BBX  (Fig.  55),  are  drawn  from  the 

y     points  A,  B,  of  the  straight  line  AB,  to 

the  vertical  straight  line  MN,  AXBX  is 

B    called    the    vertical    projection    of    AB 

'  on  MN. 

AXBX  =  A2B  =  AB  -sin  6. 

.*.  the  length  of  the  vertical  projection  of 
a  given  length,  I,  is  I  times  the  sine  of  the 
projecting  angle,  this  angle  being  meas- 
ured from  the  right-hand  horizontal  line 
as  initial  line,  the  starting-point  for  the 
length  being  taken  as  the  pole,  A  for  AB,  B  for  BA 

Similarly,  the  projection  of  ABO 
(Fig.  56)  is  lx  •  sin  $1  +  l2  •  sin  62,  or, 
the  projection  of  a  sum  is  the  sum  of 
the  projections.  Counting  projec- 
tions upward  as  plus  and  downward 
as  minus,  the  projection  of  any  closed 
outline  is  evidently  zero.  That  of 
Fig.  57  is  •        *  FlG-56- 

AXBX  +  BxOx  +  Ci©,  +  A^i  +  -#1*1  +  FXAV 
or  zero,  since  there  is  a  return  to  the  starting-point.  In 
surveying  this  is  expressed  by  saying  that  the  sum  of  the 
jy*  northings  is  the  same  as  the  sum  of 
the  southings,  in  a  closed  diagram 
D\  representing  a  plotted  survey.  The 
lengths  of  the  sides  being  lv  l2,  l3, 
.  .  .  ln,  the  sum  of  the  projections  is 

lx  •  sin  0j  -|-  l2  ■  sin  62  +  l3  •  sin  03 
+  .   .   .  +Zw.sin0n=O. 

What  is  here  termed  vertical  projec- 
tion is  identical  with  difference  of 
latitude  in  surveying,  or,  as  com- 
monly called,  latitude. 


§  86]  THE   SINE   FAMILY.  147 

.-.  Difference  of  latitude  of  two  points  is  their  distance  apart 
multiplied  by  the  sine  of  the  bearing  of  their  connecting  line 
(or  "  course  ")  from  the  east  and  west  line. 

LABORATORY  EXERCISE. 

Draw  a  line  and  its  projection  on  a  vertical  line.  Measure  the  lines 
and  the  angle.  Compare  the  measured  length  of  the  projection  with  its 
calculated  length.  Taking  the  projection  as  given,  calculate  the  length 
of  the  projected  line  and  compare  with  the  measurement. 

Carry  out  the  preceding  exercise  in  the  field  instead  of  on  paper. 

Make  the  projection  of  a  closed  broken  line  carefully  in  a  drawing, 
measure  each  projection,  and  add  algebraically.  See  how  near  the  result 
is  to  zero. 

Do  the  same  in  the  field. 

EXERCISES. 

Find  the  latitude  of  each  of  the  following  courses,  both  by  natural  sines 
and  by  logarithms,  and  test  the  results.  The  lengths  of  the  courses  are 
given  in  chains,  and  the  bearing,  or  direction  from  the  north  and  south 
line,  is  indicated  by  the  corresponding  letters.  Thus,  N  20°  E,  10  chains, 
signifies  that  the  course  is  10  chains  long,  and  is  directed  20°  toward 
the  east  from  the  north  line.  So  is  S  20°  E,  20°  east  from  the  line  running 
south  from  the  beginning  of  the  course. 

1.   N  20°  13'  W,  18.34  chains.  2.   N  34°  E,  17.35  chains. 

3.   S  54°  34'  E,  85.45  chains.  4.   S  47°  47'  W,  24.17  chains. 

5.  N  56°  56' E,  34.34  chains. 

6.  Calling  north  latitude  plus  and  south  latitude  minus,  what  is  the 
bearing  of  a  course  whose  latitude  is  -f  17.21  and  length  56.13  chains? 
What  the  bearing  when  the  latitude  is  —  19.23  and  length  75.34?  How 
many  different  courses  satisfy  these  conditions? 

7.  The  notes  of  a  survey  give  the  following  courses;  determine 
whether  the  sum  of  the  northings  is  numerically  the  same  as  the  sum 
of  the  southings.  Solve  by  natural  sines  and  by  logarithms  and  compare 
results : 

N  69°  E,  437.0  feet.  S  19°  E,  236.0  feet. 

S  27°  W,  244.0  feet.  N  71°  W,  324.0  feet. 

N  19°  W,  183.0  feet. 

8.  In  an  example  like  7,  with  the  data  on  sides  reading  to  four  signifi- 
cant figures,  are  the  angles  given  sufficiently  exact?  Should  they  not 
read  to  minutes?  Would  dropping  the  final  zeros  leave  the  indicated 
accuracy  unchanged  ? 


148  PLANE  TRIGONOMETRY.  [§87 

9.   Make  up  an  example  calling  for  the  use  of  a  five-place  table. 

10.  One  calling  for  a  six-place  table. 

11.  One  calling  for  a  seven-place  table. 

12.  If  a  velocity  along  a  line  inclined  at  an  angle  A0  to  the  hori- 
zontal is  represented  by  a  line  of  given  length,  what  line  will  represent 
the  vertical  component  of  such  velocity  ? 

13.  What  is  the  initial  vertical  velocity  of  a  projectile  which  comes 
from  a  gun  at  an  angle  6  to  the  vertical  with  a  velocity  of  V  ft.  per 
second?    What  is  such  velocity  when  V  =  26.3  and  0  =  21°? 

14.  If  a  weight  of  w  pounds  rests  (tied)  on  a  smooth  plane  inclined  at  an 
angle  6  to  the  vertical,  what  is  the  pressure  on  the  plane?  What  is  the 
pressure  when  $  =  23°,  and  the  weight  10  pounds?  What  is  the  pull  on 
a  string  held  parallel  to  the  plane  and  holding  such  a  weight  ? 

15.  A  lever  a  feet  long  and  inclined  at  an  angle  A°  to  the  vertical  has 
a  weight  of  p  pounds  at  one  end,  what  is  the  moment  of  this  weight  about 
the  other  end  of  the  lever?  What  is  it  when  p  =  10  pounds,  a  =  10  feet, 
^°  =  10°? 

§  87.     Horizontal  Projections. 

To  use  the  sine  in  horizontal  projections,  the  upright 
vertical  is  taken  as  the  initial  line  and  the  clockwise  turn  as 

positive.     Then  projections  running 

£ —*      toward  the  right  are  plus,  and  those 

toward  the  left,  minus. 

What  has  been  said  with  reference 
to  sums  of  vertical  projections  holds 
A  -  also  for  sums  of  horizontal   projec- 

tions. In  surveying,  the  horizontal 
projection  of  a  course  is  called  its  departure,  or  the  differ- 
ence in  longitude  of  its  extremities.  The  algebraic  sum  of 
the  departures  of  a  closed  survey  is  zero. 

EXERCISES. 

1.  Determine  the  departures  of  the  courses  given  in  Exs.  1-5  on 
page  147. 

2.  In  Ex.  6,  page  147,  change  the  word  "  latitude  "  to  departure,  the 
word  "  north  "  to  east,  the  word  "  south  "  to  west,  and  solve. 

3.  Determine  whether  the  sum  of  the  eastings  is  the  same  as  the  sum 
of  the  westings  in  the  survey  of  Ex.  7,  page  147. 


THE  SINE  FAMILY.  149 

EXERCISES  TO  BE  SOLVED  WITHOUT  THE  USE  OF  TABLES. 

Express  the  results  in  terms  of  the  sines  given  in  §  75,  and  in  general 
terms. 

1.  A  regular  polygon  of  n  sides  is  inscribed  in  a  circle  of  radius  r ; 
find  the  general  expression  for  the  length  of  its  sides  in  terms  of  the 
radius  and  half  the  central  angle  subtended  by  a  side.  To  what  inscribed 
polygons  may  the  expression  be  applied  without  using  the  tables? 
Determine  the  sides  of  such  polygons  when  the  radius  is  250  feet. 

2.  Determine  the  general  expression  for  the  perpendicular  from  the 
centre  upon  the  sides  of  the  polygons  of  Ex.  1,  in  terms  of  the  radius  and 
the  half  of  the  vertex  angle.  To  what  polygons  may  this  be  applied  with- 
out using  the  tables  ?    Make  the  calculations  when  the  radius  is  100  feet. 

3.  If  a  building  is  50  ft.  wide,  calculate  the  length  of  rafters  when  the 
pitch  of  the  roof  is  30°,  45°,  60°,  respectively,  the  roof  being  an  isosceles  tri- 
angle. Also  the  rise  of  the  roof  in  each  case.  If  A  is  the  pitch  and  I  the 
length  of  rafter,  what  is  the  rise  ?  What  the  width  ?  If  b  is  the  breadth 
and  A  the  pitch,  what  is  the  length  of  rafter?  What  the  rise?  If  p  is 
the  rise  and  A  the  pitch,  what  is  the  length  of  rafter?     What  the  width? 

4.  A  ladder  50  feet  long  is  tilted  at  angle  of  30°  to  the  horizontal  and 
leans  against  a  building ;  find  the  distance  of  the  top  of  the  ladder  from 
the  ground,  and  of  the  foot  from  the  building  ?  Answer  the  same  ques- 
tions when  the  tilt  is  45° ;  60°. 

5.  A  kite  is  flying  50  feet  high  ;  what  is  the  length  of  string  when  the 
angle  between  the  string  and  ground  is  30°,  assuming  the  string  straight? 
When  the  angle  is  45°?  60°? 

6.  The  upper  part  of  a  tree  broken  over  by  the  wind  makes  an  angle 
of  30°  with  the  ground,  and  the  top  lies  50  feet  from  the  root ;  how  tall 
was  the  tree  ? 

7.  The  angle  of  elevation  of  a  hill  (its  rise  from  a  horizontal  line)  is 
30°;  what  distance  would  one  travel  in  walking  up  the  hill  to  an  elevation 
of  100  feet?  What  if  the  angle  were  45°?  60°?  What  if  the  angle  were 
A  and  the  height  h  ? 

8.  Find  x,  y,  in  the  adjoining  diagram.  If 
30°,  60°  are  changed  to  A,  B,  and  10  yards  to 
a  yards,  find  expressions  for  x,  y,  in  terms  of  sines. 
(Drop  perpendicular  from  B  on  AC.) 

9.  A  regular  polygon   of  n   sides  is  circum-  A       .10  yds.     b  y 
scribed  about  a  circle  of  radius  r;  find  an  expres-  Fig.  59. 
sion  for  the  length  of  the  diameter  of  the  polygon 

in  terms  of  the  sine  of  half  the  central  angle ;  find  also  an  expression  for 


150  PLANE   TRIGONOMETRY. 

the  length  of  the  side  of  the  polygon.  Taking  the  radius  as  50  feet,  find 
the  lines  just  mentioned  for  all  regular  polygons  for  which  they  may  be 
found  without  using  the  tables. 

10.  A  person  on  the  top  of  a  tower  80  feet  high  observes  two  objects 
on  a  straight  road  in  line  with  the  tower.  The  angle  of  depression 
of  one  object  below  a  horizontal  line  passing  through  the  top  of  the 
tower  is  30°,  and  that  of  the  other  is  45°.  Find  the  distance  apart  of  the 
two  objects.  Solve  the  same  problem  when  the  angles  are  A,  B,  and 
the  height  h. 

11.  A  man  wishing  to  know  the  width  of  a  river,  selects  a  tree  on  the 
opposite  bank,  and  finds  its  angle  of  elevation  to  be  45°.  Going  back  150 
feet  in  a  straight  line  with  the  tree  and  first  point  of  observation,  the 
elevation  is  30° ;  how  wide  is  the  river  ?  Solve  the  same  problem  when 
the  angles  are  60°  and  30°;  60°  and  45°;  also  when  they  are  A,  B,  with 
the  distance  150  changed  to  d. 

12.  How  far  from  the  polar  axis  of  the  earth  is  a  point  whose  latitude 
is  30°  (the  radius  of  the  earth  being  taken  as  4000  miles)?  45°?  60°? 
Solve  the  same  problem  when  the  latitude  is  L  and  radius  r. 

13.  Find  the  area  of  an  equilateral  triangle  whose  side  is  100  feet. 
Solve  the  same  problem  when  the  side  is  s.  Find  also  the  distance  of  the 
centre  of  gravity  of  such  a  triangle  from  the  vertex  and  from  a  side. 

14.  The  hypothenuse  of  a  right-angled  triangle  is  15  yards ;  what  is 
its  area  when  one  of  its  angles  is  30°  ?  What  when  one  of  its  angles  has 
its  sine  equal  to  §  ? 

15.  The  base  of  a  right-angled  triangle  is  24  feet;  what  is  its  area 
when  one  of  its  angles  is  30°?  What  when  one  of  its  angles  has  its  sine 
equal  to  f  ? 

EXERCISES  USING  THE  TABLES. 

1.  Make  up  from  each  of  the  preceding  fifteen  examples  an  example  for 
whose  solution  it  is  necessary  to  use  the  tables,  by  taking  the  angles  other 
than  30°,  45°,  60°,  etc.  Solve  the  same,  and  apply  in  each  case  a  suitable 
test.  Solve  each  example  by  Natural  Sines  and  by  Logarithmic  Sines, 
and  compare  results.  Make  in  each  case  an  example  calling  for  the  use 
of  a  four-place  table,  and  one  calling  for  the  use  of  a  five-place  table. 
Suggest  data  suitable  for  a  six-place  table ;  for  a  seven-place  table. 

2.  A  smooth  lever,  turning  in  a  vertical  plane,  has  a  1-pound  weight 
attached  to  its  end ;  what  connection  with  a  table  of  sines  (of  the  angles 
which  the  lever  makes  with  the  vertical)  have  the  pressures  at  right 
angles  to  the  lever?    What  if  the  weight  is  10  pounds? 


88] 


THE   SINE   FAMILY. 


151 


§  88.     Solving  by  Sines  Triangles  Other  than  Right-angled 
Triangles. 

To  solve  an  oblique-angled  triangle  there  must  be  given 
not  less  than  three  parts,  including  a  side.  Only  three  dif- 
ferent cases,  so  far  as  method  of  solution  is  concerned,  can 
arise  :  % 

(a)  A  pair  of  opposites  (an  angle  and  its  opposite  side)  is 
among  the  parts  given. 

($)  The  three  sides  are  given. 

(7)  Two  sides  and  the  included  angle  are  given. 

Case  (a)  is  always  solved  by  sines.  Case  (/3)  may  be 
solved  by  sines,  though  this  is  not  the  best  solution  when  all 
three  angles  are  desired  (see  §  185).  Case  (7)  cannot  be 
solved  by  sines  conveniently  when  the  sides  are  long.  It 
can  be  solved  by  dividing  the  triangle  into  right-angled 
triangles. 

A  pair  of  opposites  are  given. 

Drop  a  perpendicular,  p,  from 
0  on  AB  (Fig.  60),  or  AB  pro- 
duced (Fig.  61).  Then  from  the 
two  triangles  formed  we  have 

p  =  a  •  sin  J?, 

p  =  b  •  sin  A, 

.\  a  •  sin  B  =  b  •  sin  A. 


Case  (a). 


sin  A  sin  B 
By  symmetry,  each  one  of  these  expressions  is  the  same  as 
.  c  .  This  may  be  proven,  if  thought  necessary,  by  drop- 
ping the  perpendicular  from  some  other  vertex  than  C.  How- 
ever, it  is  advisable  to  note  early  the  principle  of  symmetry, 
that  where  certain  parts  of  a  diagram  are  involved  symmet- 
rically in  an  expression,  the  remaining  parts  are  similarly 
involved. 


152 


PLANE   TRIGONOMETRY. 


[§88 


It  is  evidently  a  matter  of  indifference  whether  the  per- 
pendicular falls  within  the  triangle  or  without,  since  the 
c  sine  of  the  exterior  angle  of  a  triangle  is 

the  same  as  the  sine  of  the  interior  adjacent 
supplementary  angle.     Thus,  in  all  cases, 

a  b     _     c 

sin  A     sin  B     sin  C 

Each  of  these  expressions  is,  in  fact,  the 
diameter  of  the  circle  circumscribing  the  triangle. 

The  central  angle  BOO  (Fig.  63)  is  twice  the  angle  A  on 
the   circumference.      Taking    OD   perpen- 
dicular to  BO,  angle  BOD  is,  therefore,  A. 

.-.  DB  =  OB  -sin BOB; 
r  •  sin  A. 


or, 


2  r,  or  the  diameter  of  the  circle. 


Fig.  63. 


sin  .4. 

Show  how  to  construct  the  diagram  to  bring  in  the  angle 
B  or  0.     The  same  relation  holds  also,  of  course,  for  right- 
angled  triangles.     Then  0  is  the  middle  point  of  AB,  and 
a  b      _  c  _       c  c 

sin  A      sin  B     1      sin  90°      sin  0 

In  using  the  relations  above  in  solving  an  oblique  triangle, 
one  should  proceed  as  in  right-angled  triangles,  writing  on 
the  first  side  of  an  equality  the  part  desired,  then  immedi- 
ately on  the  other  side  the  part  opposite,  followed  by  the 
remaining  given  parts,  in  the  order,  sine/side  or  side/sine, 
according  as  the  first  member  is  sine  or  side,  as  in 

■,  when  J.,  B,  b  are  given. 


sin  A 


a  =  sin  A 


sin  B 

c 


sin  A  =  a 


sin  A  =  a 


sin  0 
sini? 

sin  C 


,  when  A,  C,  c  are  given. 


,  when  a,  5,  B  are  given. 
,  when  a,  <?,  C  are  given. 


88] 


THE   SINE   FAMILY. 


153 


The  student  may  write  the  corresponding  formulas  for  the 
eight  other  possible  cases. 

In  solving  triangles  by  these  formulas,  only  two  different 
cases  can  arise : 

(1)  G-iven  a  pair  of  opposites  and  another  angle, 

(2)  G-iven  a  pair  of  opposites  and  another  side. 

In  Case  (2)  an  angle  is  to  be  found  from  its  sine,  and  since 
it  is  an  angle  of  a  triangle,  and  therefore  less  than  180°,  there 
are  two  angles  which  have  the  calculated  sine  ;  for  if  x  be 
one,  180°  —  x  is  another.  It  is  necessary  in  all  cases  to  deter- 
mine which  angle  to  take,  or  whether  both  are  to  be  taken. 

f  >  90°       (i) 
In  Case  (2)  the  given  angle  is   |  <  g()</      \/ 

In  Case  (i)  each  of  the  remaining  angles  must  be  less  than 
90°,  and  there  is  no  ambiguity. 

In  Case  (ii)  there  may  arise  four  different  cases : 

(8)  The  side  of  the  given  pair  of  oppo- 
sites is  longer  than  the  other  given  side. 
For  example,  given  a,  A,  b,  with  a>  b. 
Since,  by  hypothesis,  A  is  acute,  so  is  B, 
for  if  a  >  b,  so  is  A  >  B. 

(e)  The  side  of  the  given  pair  of  oppo- 
sites is  less  than  the  other  given  side. 

Under  this  there  may  fall  three  sub- 
cases, as  in  the  following  diagrams : 

(ex)  In  Fig.  65,  the  side  opposite  the 
given  angle  is  not  long  enough  to  reach 
to  the  border  of  the  angle.     Here 

a  <  b  and  also  <  b  •  sin  A, 

and  there  is  no  triangle. 

(e2)  In  Fig.  66,  a  is  just  long  enough  to 
reach  AB.     Here 

a  <b,  but  a  =  b  •  sin  A, 

and  there  is  only  one  triangle. 


Fig.  65. 


Fig.  66. 


154  PLANE   TRIGONOMETRY.  [§88 

(e3)  In  Fig.  67  the  side  a  is  longer 
than  just  necessary  to  reach  the  base,  so 
that  it  may  rest  in  two  positions,  CBV 
OB2.     Here 

a  <  6,  but  >  b  -  sin  A. 

This  is  the  only  case,  when  a  pair  of  opposites  are  given, 
in  which  both  possible  angles  determined  by  the  sine  are  to 
be  taken.     It  is  known  as  the  ambiguous  case. 

In  solving  problems  where  a  pair  of  opposites  and  another 
side  are  given,  one  should  note,  therefore : 

(f )  If  the  given  angle  is  greater  than  90°,  the  calculated 
angle  is  to  be  taken  less  than  90°. 

(77)  If  the  given  angle  is  less  than  90°,  with  the  opposite 
side  greater  than  the  other  given  side,  the  calculated  angle  is 
to  be  taken  less  than  90°. 

(/c)  If  the  given  angle  is  less  than  90°,  with  its  opposite 
side  less  than  the  other  given  side,  there  may  be  no  triangle, 
one  triangle,  two  triangles. 

Which  of  the  three  cases  of  Case  (/e)  is  present,  may  be 
settled  by  making  a  preliminary  sketch  to  scale  of  the  data, 
with  the  results  already  indicated  in  (ej),  (e2),  (e3).  Lay 
out  the  given  angle  with  the  protractor.  From  its  vertex, 
lay  out  along  one  of  its  terminals  the  side  bordering  the 
angle.  From  the  other  extremity 
of  this  side  as  a  centre,  describe 
a    circle    having    for    its    radius 

the  side  opposite  the  given  angle.     //        x  N-^"~    "'^°V 
There  will  be  either  no  triangle, 
or  one  triangle,  or  two  triangles 

(Fig.  68).  A  preliminary  sketch  is,  however,  not  essential. 
The  calculation  of  the  unknown  parts  will  itself  show  which 
case  is  present. 

For  example,  if  a,  A,  5,  are  given,  with  A  <  90°, 

.     t>      1  sin  A 

sm  B  —  0 


§88] 


THE   SINE   FAMILY. 


155 


•  (X)  If  a  <  b  -  sin  A,  sin  B  will  be  greater  than  1,  showing 
there  is  no  triangle.  Using  logarithms,  log  sin  B  would 
appear  greater  than  zero. 

(/*)  If  a—b  •  sin  J.,  sin  jB=1,  and  there  is  but  one  triangle, 
with  B  =  90°.     Using  logarithms,  log  sin  B  will  be  zero. 

(v)  If  a  >  b  •  sin  A,  sin  B  <  1,  and  there  are  two  solu- 
tions.    Using  logarithms,  log  sin  B  will  appear  less  than 

zero. 

Model  Examples. 

c 


1.   Given 


'  a  =  65 
4  =  73° 
.6  =  97 


sin  B  =  b 


Find  <  C    No  triangle. 


C  =180  -(A  +  B) 
c  =  sin  0 


The  order  here  is  that 
in  which  the  quan- 
tities will  be  found, 
as  that  also  after 
"Find"  above. 


sin  A 

log  sin  B  =  log  b  +  log  sin  A  —  log  a 
log  b  =  log  a  +  log  sin  A 
log  c  =  log  sin  0  +  log  a  —  log  sin  A. 

SOLUTION  BY   LOGARITHMS. 

log  5  =  1.9868 

log  a  =  1.8129 

log  b  -log  a  =  0.1739 

log  sin  A  =  1.9806 


SOLUTION   BY   SINES. 

sin  A  =  0.9563 
5=97. 
86.0 
6.7 


65  [93 
sin  >  1 
no  triangle. 


log  sin  B=  0.1545 

This  is  >  0 
.\  no  triangle. 


156 


J  LANE   TRIGONOMETRY. 


[§88 


f  a  =  1072 
2.   Given]  A=12°W 

{  6  =  4849     * 

sin  B  =  b 


-N90 


Fig.  70. 


r£  =  90° 
Find]  C  =  77°14' 
I  c=4731 


sin  A 
a 


(7=180-  Oi  +  £) 
c  =  sin  C 


sin  ^4 
log  sin  B  =  log  b  —  log  a  +  log  sin  A 

log  e  a  log  sin  C  +  log  a  —  log  sin  J.. 


SOLUTION   BY  NATURAL   SINES. 

5  =  4849. 
sin  .4  =        0.2210 
969.8 
97.0 
4.8 


1072,  about 
sin  B  =1,  about. 


SOLUTION    BY   LOGARITHMS. 

log  b  =  3.6857 

loga=  3.0302 

log  b  —  log  •  a  =  0.6555 

log  sin  A  =  1.3444 

log  sin  #=1.9999 

log  sin  B  =  0,  about. 


Thus  the  triangle  is  seemingly  a  right-angled  triangle. 

In  determining  <?,  one  should  not  set  down  again  any  logs 
already  set  down.  Both  B  and  e  should  be  determined  in 
the  same  scheme,  as  follows  : 

log  b  =  3.6857  (1) 

log  a  =  3.0302  (2) 

log  6 -log  a  =0.6555  (4) 

log  sin  4  =  1.3444  (3) 

log  sin  #  =  1.9999  (5) 

log  a  -  log  sin  A  =  3.6858  (6) 

log  sin  C=  1.9891  (7) 

log  em  3.6749  (8) 

The  corresponding  values  of  B,  c  are  entered  after  "Find" 
above. 


88] 


THE   SINE   FAMILY. 


157 


A  triangle  like  the  preceding  can  occur  only  by  design. 
Whether  the  triangle  is  exactly  a  right-angled  triangle,  can- 
not be  settled  by  the  tables.  It  might  appear  to  be  right- 
angled  by  a  four-place  table,  where  a  seven-place  table  would 
show  otherwise.  No  table  can  settle  the  matter  except 
within  its  own  degree  of  approximation. 

Testing  the  preceding  results :  A  combination  of  the  parts 
in  some  new  form  is  a  test.     Here  we  may  use 

c  •  sin  A  =  a  •  sin  (7,  or  log  •  c+log  .  sin  A  =  log  •  a-f-log  •  sin  0. 

The  logs  are  to  he  looked  up  in  the  tables,  and  not  taken  from 
the  preceding  scheme,  in  order  to  test  the  quantities  found. 


log  a 
log  sin  C 


3.0302 
1.9891 
3.0193 


log<?  =  3.6749 

log  sin  A  =1.3444 

3.0193 


The  sums  agree.     They  might  have  differed  by  2  in  the 
final  figure  and  the  solution  be  considered  correct  (§  82). 


[a   =52.1 
3.   Given  -l  A  =  31°  25' 
[b    =61.2 


Find 


B  =  37°  45' 
or  142°  15' 
C  =  110°  50' 
or  6°2(y 
c  =93.4 
or      11.0. 


Fio.  71. 


sin  B  —  b  - 


sin  A 


C=180°-(^  +  JB) 


c  =  sin  (J - 

sin  A 

log  sin  B  =  log  b  +  log 


sin  A 


log  c  =  log  sin  C+  log  -: — -• 
sin  A 


158 


PLANE  TRIGONOMETRY. 


[§88 


SOLUTION   BY   NATURAL    SINES. 

sin^L  =    0.521 
6  =  61.2 
31.26 
.52 
.10 
81.9  =  b  sin  A 
0. 612  =  sin  B 
521 | 319 
313 
6 
5 
1 

sin  B  <  1 ;  .*.  two  i?'s. 

sinCi=   0.935 
a  =  52.1 
46.75 
1.87 
sin  A  .09         cx 

0.5215148.71|93.4 
46.93 
1.78 
1.56 
.22 
.20 
sin(72=    0.110 
0  =  52.1 
5.50 
.22 
sin  A  .01         c2 

0.522|5.73|11.0 
5.22 
.51 
.52 


SOLUTION   BY   LOGARITHMS. 


LOGARITHMS. 

sin  B=  1.7870  < 

^6)  two  .S's. 

6=1.7868 

(3) 

sin  ^  =  2.0002 
a 

(4) 

sin  .4=1.7170 

(1) 

0  =  1.7168 

(2) 

a     =1.9998 
sin  A 

(5) 

sin  (72  =  1.0426 

0) 

nn{7r=  1.9706 

(8) 

c2  =  1.0424 

(9) 

^  =  1.9704 

(10) 

Check:  b  sin (7= 

=  c  sin  i?. 

log  6  = 

=  1.7868 

log  sin  Cx  = 

=  1.9706 

1.7574 

log*lS 

=  1.9703 

log  sin  B  = 

=  1.7869 

1.7572 

This  log  check  is  required 
to  check  to  within  2  in  the 
third  place,  since  the  lines 
of  data  are  three-figured.* 


*  Example  3  is  taken,  as  to  data, 
from  a  book  on  Trigonometry,  ex- 
cept that  I  have  made  the  angle 
read  25'  instead  of  23',  since  three- 
figured  data  in  lines  call  for  angles 
reading  not  closer  than  to  the  near- 
est five  minutes,  as  previously  ex- 
plained.   In  the  book  referred  to, 


88]  THE   SINE  FAMILY.  159 


4.  Given 


a  =  7.42 

A  =  105°  15'  Find 

b  =  3.39 


Since  A  >  90°,  with  a  >  5,  there  is  one  and  but  one  solu- 
tion. The  student  may  make  the  calculations,  using  both 
natural  sines  and  logs,  determining  also  how  far  the  signifi- 
cant figures  of  the  calculated  results  should  be  taken. 

5.  The  data  as  in  Ex.  4,  with  the  values  of  a,  b,  interchanged. 

The  solution  is  impossible,  since  if  a  <  6,  then  is  A  <  B. 
As  A  is  obtuse,  this  would  require  B  to  be  obtuse.  A  tri- 
angle cannot  have  two  obtuse  angles. 


6.   Given 

1 

Find 

a  =  20.24 

\ 
\ 

C  =  57°  3' 

A  =  103°  36' 

\103O36' 

\ 

19'21?^ 

b  =  6.901 

B  =  19°  21' 

A 

Fig.  72. 

B 

c  =  17.48 

C=  180° 

-(A 

+  tf) 

,_sini?. 

a 

C  - 

sin  C  -  a 

sin  A  sin  A 


Order  of  logs  { log  b  =  log  a  ~  log  sin  A  +  log  sin  B 

I  log  c  =  log  a  —  log  sin  ^4  -f-  log  sin  O 

Since  A  >  90°,  with  a  >  J,  there  is  only  one  solution. 


this  example  is  worked  out  completely  as  a  "model  example."  The  calcu- 
lated sides  are  carried  to  seven  significant  figures,  and  the  calculated  angles 
to  hundredths  of  a  second.  A  procedure  like  that  is  justified  only  in  the 
case  of  data  known  exactly,  which  is  never  the  case  with  any  triangle  met 
"on  land  or  sea."  When  one  is  called  upon  to  measure  a  line  and  finds 
himself  satisfied  on  knowing  the  length  to  within  one  five-hundredth  of  its 
value,  the  spirit  of  calculation  in,  him  seems  somewhat  insatiable  if  nothing 
short  of  pushing  the  apparent  error  in  results  to  within  one  ten-millionth  of 
the  calculated  line  meets  his  wishes.  This  refinement  of  calculation  is  all  a 
sham.     The  figures  beyond  the  place  of  the  data  are  altogether  worthless. 


160 


PLANE   TRIGONOMETRY. 


[§88 


SOLUTION   BY   NATURAL    SINES. 


sin  B  = 

0.3314 

a  — 

20.24 
6.628 
.066 

sin  A 

.013 

b 

0.9720 

6.707 

6.900 

5.832 

'    .875 

.875 

sin  (7  = 

0.8392 

a  = 

20.24 
16.784 

.168 

sin  A 

.033 

c 

0.9720 

16.985 
9.720 

17.48 

7.265 

6.804 

.461 

.389 

SOLUTION   BY   LOGARITHMS. 

log  a  =  1.3062 
log  sin  A=  1.9876 
diff.  =1.3186 

log  sin  #  =  1.5203 
log  sin  0=  1.9238 
log  b  =  0.8389 
log  o         =  1.2424 


(The  student  may  supply  a 
log-check.) 


.072 


§  89.     Schemes  for  Solution. 

The  preceding  examples  have  been  worked  out  to  serve  as 
models  for  subsequent  solutions  of  the  same  sort.  The  stu- 
dent will  note  the  forms  followed  in  the  calculations.  He 
will  state  in  tabular  form  the  parts  given,  followed  by  a  tabu- 
lar statement  of  the  parts  to  be  found,  in  the  order  in  which 
they  will  be  found,  accompanied  by  a  diagram  to  scale  on 
which  the  given  parts  are  noted  in  full  lines,  with  the  lines 
to  be  calculated  indicated  by  dotted  lines.  As  the  parts  cal- 
culated are  determined,  they  may  be  entered  in  the  table  for 
such  parts.  The  formulas  to  be  used  in  calculating  the  parts 
are  to  follow  the  data  as  in  the  examples  above,  and  when 


89] 


THE   SINE   FAMILY. 


161 


logs  are  to  be  used,  the  order  in  which  they  are  to  follow 
each  other  in  the  scheme  of  solution  may  be  indicated  as 
above,  while  the  student  is  still  in  the  novitiate  stage. 

The  order  in  which  the  logs  follow  each  other  in  the  solu- 
tion scheme  should  be  such  as  to  make  the  manipulations  easy. 
Some  regular  scheme  ought  always  to  be  followed.  That 
adopted  in  the  preceding  examples  seems  as  economical  as  any. 

Having  settled  on  a  scheme,  enter  on  the  scheme  all  logs 
which  are  called  for  before  manipulating  these  logs  at  all. 

(A)  The  complete  scheme,  using  logs  (and  not  cologs) 
when  a  pair  of  opposites  and  another  side  are  given  (assum- 
ing only  one  solution),  is  for  a,  A,  b  as  follows : 

c 

ml  Am  FindJ(7  = 

I  b  = 


l  c 


Fig.  73. 


Formulas 


sin  B  =  b  • 


sin  A 


C=1$Q°-(A  +  B) 
c  =  sin  C' 


Log  order 


sin  A 
log  sin  B  =  log  b  +  log  f^A\ 

log  c  =  log  sin  C  +  log  ( ) 

Vsin  A) 

Log  Solution  Scheme. 


(i) 

(2) 
(3) 


LOGARITHMS. 

sin  B  = 

(6) 

b  = 

(3) 

sin  A 
a 

(4) 

sin  A  = 

(1) 

a  = 

(2) 

a 

(5) 

sin  A 

sin  0  = 

(7) 

c  = 

(8) 

162  PLANE   TRIGONOMETRY.  [§89 

In  working  examples,  make  the  scheme  first.  Then  fill  it 
in.  First  with  (1)  ;  then  (2),  (3)  ;  subtract  (1)  from  (2) 
for  (4) ;  then  (2)  from  (1)  for  (5) ;  add  (3),  (4),  for  (6). 
Then  find  B.  Then  0.  Next  fill  in  (7)  ;  add  (5),  (7),  for 
(8).     Then  look  up  c. 

TESTING   RESULTS. 

The  student  should  test  his  own  results.  Combining  the 
logs  in  some  new  relation  is  an  appropriate  test  here,,  as  in 

b  .  sin  0  =  e  •  sin  B. 

logc  = 


log  b  = 
log  sin  C  = 


log  sin  B 


Make  sure  to  look  up  all  logs  (of  "Ans.")  in  the  tables. 
The  log-check  should  check  by  2  or  less  in  the  final  figure, 
the  final  figure  being  the  figure  in  the  place  to-which  the  data 
go.  For  two-figured  data,  the  second  figure  of  mantissa  is 
the  final  place,  etc.     Test  also  by  making  diagrams  to  scale. 

(i?)  Solution  scheme  for  case  (J.)  when  cologarithms  are 
used: 

log  sin  B  =  (6) 

log  6  =  (3) 

colog  a  =  (4) 

log  sin  A  =  (1) 

colog  sin  A  =  (5) 

loga=  (2) 

log  sin  0=  (7) 

log<;=  (8) 

The  numerals  indicate  the  ordering  of  performing  the 
work:  (1),  (4),  (3)  added  give  (6).  Then  find  B.  Then 
C.  Then  (7).  Then  add  (5),  (2),  (7),  for  (8).  Then 
find  c. 

EXERCISES. 

Arrange  calculation  schemes  for  all  similar  cases,  and  the  logarithmic 
tests.     Arrange  similar  schemes  when  cologarithms  are  used. 


89] 


THE   SINE   FAMILY. 


163 


Solve  the  following  examples  both  by  logarithms  and  by  natural  sines. 
Compare  results.  Give  also  the  logarithmic  tests.  Make  diagrams  to 
scale.  Measure  the  ungiven  parts,  and  compare  with  the  results  of  cal- 
culation. Make  sure  the  results  are  given  appropriately  as  to  significant 
figures  and  angle  readings. 


1.  a8,A3o°,bS. 

2.  a  05,  A  47°,  b  35. 

3.  a  5.4,  A  67°  30',  b  6.7. 

4.  a  71.5,  A  65°,  b  63.2. 


6.  a  67.34,  ^  45°  43',  6  87.34. 

7.  a  6715,  ,4  63°  13',  6  7123. 

8.  a  23.579,  A  34°  52'  23",  b  25.431. 

9.  a  234.671,  A  34°  45'  67".3,  b  234.678. 


5.   a  7.45,  ^47°  25',  6  9.36.    10.   a  2345.76,  ,4  56°  57'  58".45,  b  2531.89. 

11.  Construct  similar  examples  for  a,  A,  c,  taking  data  to  one  signifi- 
cant figure  in  lines;  two  significant  figures ;  three;  four;  five;  six;  seven. 
See  that  the  angles  read  appropriately.     Solve  and  test  them. 

12.  Do  the  same  for  6,  B,  c,  as  given. 

13.  Do  the  same  for  C,  c,  b,  as  given. 

14.  Construct  some  examples  which  have  no  solution.     Test. 

15.  Construct  some  examples  with  two  solutions.     Solve. 

16.  Construct  some  examples  with  only  one  solution.     Solve. 

((7)  The  complete  scheme  when  a  pair  of  opposites  and 
another  angle  are  given,  is  similar  to  the  following,  assuming 
a,  A,  B  given  : 

A 

b  = 


Given 


a  = 
A  = 
B  = 

A 

B                                   C 
Fig.  74. 

r  (7=  180°  —  (A  +  B) 

Find 

• 

b  =  sin  B  •  - — - 
sin  A 

c  =  sm  0  •  - — - 

sin  A        J 

• 

c  = 


Log  order 


log  b  =  log  a  —  log  sin  A  +  log  sin  B 
log  c  =  log  a  —  log  sin  A  +  log  sin  0 


164 


PLANE   TRIGONOMETRY. 


[§89 


LOG    SOLUTION    SCHEME. 


log  b  = 
log  sin  B  = 

\sm  A) 
log  sin  A  = 

log  a  — 

log  sin  0  = 
logo  = 


(7) 
(3) 

(5) 

(2) 
(1) 

(6) 

(4) 

(8) 


SOLUTION   TEST. 

b  •  sin  Q  —  c  -  sin  i?,  or 
log  b  +  log  sin  C  =  log  <?  +  log  sin  B 
log  b  =  log  <?  = 

log  sin  0  =  log  sin  B  — 

sum  =  sum  = 

The  solution  scheme  is  arranged  from  its  centre,  (6)  being 
(5)  repeated,  to  bring  it  adjacent  to  log  sin  O  and  log  sin  B. 
In  extended  calculations,  where  the  same  log  is  to  be  added 
(subtracted)  several  times,  it  may  be  written  on  the  edge  of 
a  card  and  held  adjacent  to  the  log  with  which  it  is  to  be 
combined.  On  which  edge  it  should  be  written,  or  whether 
on  both,  will  suggest  itself.  The  card  while  being  used  is 
not  to  cover  the  part  of  the  scheme  on  which  you  wish  to 
write. 

EXERCISES. 

Construct  the  corresponding  schemes  for  all  similar  cases.     Also  the 
schemes  when  cologs  are  used. 
Solve  the  following  triangles : 

1.  a  8,  A  55°,  B  50°.  3.  a  43.1,  A  47°,  B  76°. 

2.  a  3.1,  A  50°,  B  54°.  4.  a  451,  A  59°  15',  B  58°  25'. 

5.   a  46.76,  A  67°  54',  B  68°  32'. 


§89]  THE   SINE   FAMILY.  165 

6.  a  341.67,  A  56°  32'  31",  B  54°  54'  54". 

7.  a  3214.51,  ^4  51°  51'  51".2,  B  65°  56'  56".3. 

8.  a  2345.56,  A  68°  56'  32'M2,  £  36°  36'  36".36. 

9.  Construct  similar  examples  for  other  possible  similar  cases,  seeing 
that  the  angles  read  correctly,  to  suit  the  line  data. 

GENERAL  EXERCISES  ON  THE  SINE. 

Construct  diagrams  illustrating  the  following  problems  and  solve  both 
by  natural  sines  and  by  logs.  Devise  also  a  suitable  independent  test  for 
each  numerical  example. 

1.  To  determine  the  height  of  an  inaccessible  vertical  tower,  its  angle 
of  elevation  at  two  points,  A,  B,  in  the  same  horizontal  line  as  the  foot  of 
the  tower  is  measured.  Calling  these  angles  a,  (3,  show  that  if  a  is  the 
distance  between  A  and  B,  the  height  of  the  tower  is 

sin  a  sin  /? 
sin()3~a) 

2.  A  tree  85  feet  tall  subtends  an  angle  of  58°  10'  at  a  point  A  on  the 
river's  edge.  At  a  point  B  on  the  other  side  of  the  river,  in  a  horizontal 
straight  line  with  A  and  the  foot  of  the  tree,  the  angle  of  elevation 
is  13°  50'.     Calculate  the  width  of  the  river  along  the  line. 

3.  At  a  point  on  a  horizontal  plane  the  angle  of  elevation  of  a  moun- 
tain is  34°  10',  and  at  another  point  on  the  plane,  a  mile  distant  from  the 
first  and  in  the  same  plane  of  elevation,  the  mountain's  angle  of  eleva- 
tion is  13°  30'.     Find  the  height  of  the  mountain. 

4.  From  the  top  of  a  tree,  77  feet  tall,  the  angle  of  depression  of  an 
object  on  one  bank  of  a  river  is  31°  20',  and  the  angle  of  depression  of  an 
object  on  the  opposite  bank  is  20°  10'.  Assuming  the  foot  of  the  tree  and 
the  two  objects  as  in  the  same  horizontal  straight  line,  directly  across  the 
river,  how  broad  is  the  river? 

5.  A  person  in  a  balloon  observes  the  angle  of  depression  of  the  camp 
of  an  army  to  be  about  40°.  After  rising  about  500  feet  higher,  vertically 
over  the  first  place  of  observation,  the  angle  of  depression  is  about  45° ; 
about  how  far  away  is  the  camp  ? 

6.  AB  is  a  vertical  object,  and  A,  Z),  C,  are  three  points  in  the  same 
horizontal  plane,  but  not  in  a  straight  line.  These  are  measured  angles, 
BDA  as  <j>,  BDC  as  0,  BCD  as  /?;  prove, 

AB  =  CDsin<l>sinl3~ 
sin  (/?  +  <£) 


166  PLANE   TRIGONOMETRY. 

7.  At  each  end  of  a  horizontal  base  of  length  2  a  it  is  found  that  the 
angular  height  of  a  certain  peak  is  0  and  that  at  the  middle  point  of  the 
base  it  is  <f>.    Prove  that  the  vertical  height  of  the  peak  is 

a  sin  $  sin  <fr 

V(sin  <f>  —  sin  6)  (sin  <j>  +  sin  6) 

8.  A  flagstaff,  NP,  stands  on  level  ground.  A  base,  A  B,  is  measured 
at  right  angles  to  AN,  the  points  A,  B,  N  being  in  the  same  horizontal 
plane.  The  angles  PAN,  PBN  are  6,  <f>  respectively.  Prove  that  the 
height  of  the  flagstaff  is 

a  t> sin  6  sin  <j> 

(sin  6  —  sin  <f>)  (sin  6  +  sin  <£) 

9.  Solve  Ex.  7  for  0,  24°  23';  <£,  34°  56';  a,  317.4. 

10.  Solve  Ex.  8  for  0,  67°  45';  <f>,  34°  24';  AB,  213.3. 

11.  A  man  standing  due  south  of  a  tower  on  a  horizontal  plane 
observes  the  elevation  of  the  top  of  the  tower  to  be  50°  45'.  Going  100 
yards  due  east,  he  finds  the  elevation  to  be  46°  25'.  Find  the  height  of 
the  tower. 

12.  A  man  in  a  balloon  observes  the  angle  of  depression  of  a  ship  at 
anchor  due  north  of  him  to  be  40°.  After  the  balloon  has  drifted  3  miles 
due  west  at  the  same  elevation,  the  angle  of  depression  of  the  ship  is  30°. 
Find  the  height  of  the  balloon. 

13.  From  the  extremities  of  a  horizontal  base,  AB,  whose  length  is  b, 
the  angles  BAF,  ABF  are  measured,  F  being  the  foot  of  a  tower  and  in 
the  same  horizontal  plane  as  A,  B.  At  A  the  angle  of  elevation  of  the 
tower  is  also  measured.  Call  this  a;  BAF,  0;  ABF,  <j>.  Show  that  the 
height  of  the  tower  is 

7 sin  <f>  •  sin  a > 

sin  (90° -«)  sin  (#  +  <£)' 

14.  A  vertical  object  stands  on  a  hill  whose  angle  of  slope  is  A°.  At 
a  distance  a  from  the  foot  of  the  object,  this  distance  being  measured 
along  the  slope  of  the  hill  and  down  the  hill,  the  object  subtends  an 
angle  B°.    Show  that  the  height  of  the  object  is 

sin  B 


sin  (90°  -  A  -  B) 


15.  A  house  stands  on  a  hill  whose  slope  is  15°,  and  at  a  point  80  feet 
from  the  house  down  the  hill  the  house  subtends  an  angle  of  33°.  Find 
the  height  of  the  house,  and  the  distance  from  the  point  of  observation 
to  the  top  of  the  house. 


THE   SINE   FAMILY.  167 

16.  A  vertical  object  stands  on  a  hill  whose  angle  of  slope  to  the 
horizon  is  A°.  At  two  points,  whose  distance  apart  measured  along 
the  slope  of  the  hill  is  a,  the  object  subtends  angles  B°,  C°.  Show  that 
the  height  of  the  object  is 

sin  B  sin  C 

sin  (C  -  B)  sin  (90°  +  A)' 

17.  Solve  Ex.  16  when  A  =  20°,  B  =  13°,  C  =  19°,  a  =  553  feet. 

18.  A  person  standing  on  the  slope  of  a  hill  whose  angle  of  slope  to 
the  horizon  is  A,  finds  that  the  line  running  up  hill  to  the  top  of  a  build- 
ing from  the  point  where  he  stands  makes  an  angle  B  with  the  face  of 
the  hill.  The  observer  is  a  feet  from  the  building,  this  distance  being 
measured  along  the  slope  of  the  hill  ;  show  that  the  height  of  the  build- 
ing is  sin£ 

tt  sin  (90°-  A-B)' 

19.  Solve  Ex.  18  when  a  is  75,  A  is  18°,  and  B  is  35°. 

20.  By  means  of  the  relation  that  the  sines  of  the  angles  of  triangle 
are  to  each  other  as  the  opposite  sides,  prove  that  if  a  line  is  drawn 
bisecting  one  angle  of  a  triangle,  it  divides  the  opposite  side  into  seg- 
ments proportional  to  the  sides  about  the  bisected  angle. 

21.  A  regular  pyramid  on  a  square  base  has  an  edge  150  feet  long, 
and  the  length  of  the  side  of  the  base  is  200  feet.  Find  the  sine  of  the 
inclination  of  the  face  of  the  pyramid  to  the  base,  and  the  corresponding 
angle. 

22.  A  pyramid  has  a  square  base,  the  length  of  whose  side  is  a.  The 
vertex  of  the  pyramid  is  over  the  centre  of  the  base,  and  at  a  distance  h 
from  this  centre.  Show  that  the  angle  between  the  two  lateral  faces  of 
the  pyramid  is  given  by  the  equation 

a2  +  4  h* 
Find  also  the  sine  of  the  angle  between  a  lateral  face  and  the  base. 

23.  A  flagstaff  100  feet  high  stands  in  the  centre  of  an  equilateral 
triangle.  From  the  top  of  the  flagstaff  each  side  of  the  triangle  subtends 
an  angle  of  60°.  The  triangle  is  horizontal  and  the  flagstaff  vertical. 
Show  that  the  length  of  the  side  of  the  triangle  is  50  V6  feet. 

24.  A  rectangular  target  faces  south,  being  vertical  and  standing  on 
a  horizontal  plane.  If  the  sun's  angular  altitude  is  A°  while  the  sun  is 
5°  from  the  south,  show  that  the  area  of  the  shadow  of  the  target  on 
the  horizontal  plane  is  found  by  multiplying  the  area  of  the  target  by 

sin  (90° -4)  sin  (90° -B) 
sin^. 


168  PLANE   TRIGONOMETRY. 

25.  The  angles  of  elevation  of  the  top  of  a  tower,  standing  on  a  hori- 
zontal plane,  from  two  points  distant  a,  b,  from  the  foot  of  the  tower  and 
in  a  straight  line  with  the  foot  of  the  tower,  are  together  equal  to  90°. 
Prove  that  the  height  of  the  tower  is  a  mean  proportional  between  a 
and  b.    Prove  also  that  the  sine  of  the  angle  subtended  at  the  top  of  the 

tower  by  the  line  joining  the  two  points  is  a  ~    . 

a  +  b 

26.  A  cloud  observed  simultaneously  from  two  points  on  a  north  and 
south  line  and  distant  a  miles  apart,  appears  in  the  south  from  one 
point  and  at  an  elevation  of  A,  and  in  the  north  from  the  other  point 
and  at  an  elevation  of  B ;  find  the  distance  of  the  cloud  from  each  point, 
and  its  vertical  height. 

27.  A  man  ascends  a  mountain  by  a  direct  course,  the  inclination  of 
his  path  to  the  horizon  being  at  first  a  and  afterward  changing  suddenly 
to  (3,  which  continues  to  the  top.  If  the  height  of  the  mountain  is  a 
feet,  and  the  angle  of  depression  of  the  starting-point  as  observed  from 
the  top  is  y,  show  that  the  length  of  the  ascent  is 


j  sin  (/J  -  y)  +  sin  (? -«)}-—! 


sin  y  t  j  sin  (/?  —  a) 

28.  At  noon  a  person  on  a  cliff  h  feet  above  sea  level  observes  the 

altitude  of  a  cloud  in  the  plain  of  the  meridian  to  be  a,  and  the  angle  of 

depression  of  its  shadow  to  be  /?.     If  y  is  the  angular  elevation  of  the 

sun  at  the  time  of  observation,  show  that  the  height  of  the  cloud  above 

sea  level  is  .     .        nx 

^  sin  y  sin  (a  +  /?) 

sin/3  sin  (y±«)  ' 

the  plus  sign  preceding  a  when,  if  the  cloud  is  in  the  south,  the  sun  is  in 
the  north ;  the  minus  holding  when  both  sun  and  cloud  are  on  the  same 
side  of  the  observer. 

29.  Two  circles  of  radii  a,  b,  touch  each  other  externally ;  find  ex- 
pressions for  the  sines  of  the  half  angles  between  all  possible  pairs  of 
common  tangents. 

30.  A  ship's  observer  notices  that  two  objects  are  in  a  straight  line 
whose  bearing  from  the  north  is  A°  toward  the  east.  The  ship  sails  a 
miles  in  a  course  B°  west  from  north,  and  then  the  bearing  of  the  more 
distant  object  is  D°  east  by  north  from  the  direction  of  the  ship's  course, 
and  that  of  the  other  object  is  C°  east  by  north  from  the  direction  of  the 
ship's  course ;  show  that  the  distance  between  the  objects  is 

sin  (A  +  B)  sin  (C  -  D) 
sin (C  -B  -  I)  sin (D  -  A  -  B)' 


§  90]  THE   SINE   FAMILY.  169 

31.   Show  that  if,  in  the  preceding  problem,  the  bearing  of  the  line  of 

objects  is  15°  to  the  east  of  north,  and  the  ship's  course  northwest,  and 

the  bearings  of  the  objects  east  and  northeast,  the  distance  between  the 

objects  is  sin  60°  sin  45° 

a 


sin  75°  sin  30° 


32.  A  ship  observed  another  ship  A°  from  the  north,  sailing  in  a  direc- 
tion parallel  to  its  own.  Inp  hours  its  bearing  was  B°  from  the  north, 
and  in  q  hours  afterward  its  bearing  was  C°  from  the  north.  Show  that 
if  D°  is  the  bearing  from  the  north  of  the  ships'  courses,  the  following 
equation  holds :  sin  (D  -  C)  _  p  .  sin  (B  -  C) 

sin  (D-B)~  q-  sin  (A  -  B)' 

§  90.    Use  of  the  Sine  in  Calculating  the  Areas  of  Triangles. 

Drop  a  perpendicular  from  any  vertex,  as  (7,  on  the  oppo- 
site side.     This  perpendicular  is  a  •  sin  B, 
or  b  •  sin  A,  in  all  cases.      The  double  area 
of  the  triangle  is  base  times  altitude. 

.\  2'A  =  ac-sin£,  (1) 

=  be  •  sin  A,  (2) 

=  ab  -  sin  (7,  by  symmetry.     (3) 

The  area  of  a  triangle  is  one-half  the  product  of  any  pair 
of  its  sides  and  the  sine  of  their  included  angle. 

We  have  previously  shown  (§  88)  that 

a  b  c 


sin  A     sin  B     sin  (f 
abc  abc  abc  abc 


(4) 


be  sin  A     ae  sin  B     ab  sin  C     2  A 
Thus  each  expression  in  (4)  is  — - 

7 

,\  diameter  of  circumscribing  circle  =  —  (§  88). 

...   2A=      ahc 

diameter 

By  means  of  (4),  the  area  expressions  (1),  (2),  (3),  may 
be  changed  so  as  to  involve  only  one  side  and  the  three 
angles. 


170  PLANE   TRIGONOMETRY.  [§90 

For  example,  by  (4),  c  =  sin  0  -  - — -•    Setting  this  in  (1), 

2 

Double  area  =  - — -  •  sin  B  •  sin  O 

smA 


b2 


sini? 
sin  0 


sin  C  •  sin  A,  by  symmetry 
sin  A  •  sin  i?,  by  symmetry. 


The  student  may  express  these  relations  in  a  single  verbal 
statement. 

EXERCISES. 

1.   Find  the  areas,  the  three  altitudes,  and  diameters  of  the  circum- 
scribing circles  for  the  following  triangles,  without  using  the  tables : 


a  =  14,  b  =  10,  C  =  30° 
6  =  15,  c  =  12,  A  =  120° 
a  =  15,   c  =  17,    B  =  135° 


a  =  3,  b  =  6,  C  =  60° 
a  =  6,  c  =  5,  5  =  150° 
a  =  3,    6=    5,    C  =  180° 


a  =  21,   e  =  12,    5=    45°;     a  =  3,  A  =  30°,  J5  =    30° 
ft  =  21,  ^  =  B  =  C. 

2.  Calculate  by  natural  sines  and  by  logs  the  areas  of  the  following 
triangles  and  the  diameters  of  circles  circumscribing  them.     Test. 

a  =  23.4,  b  =  34.5,  C  =  34°  25' ;  b  =  3423,  c  =  2145,  A  =  34°  23' 
a  =  3.45,  c  =  5.43,  B  =  45° 45';  a=  231,  &=  432,  C  =  23°45' 
a  =  23.4,  ^  =  45°  25',  B  =  67°  35' ;  a  =  23.1,  A  =  32°  25',  C  =  43°  35' 
c  =  6.45,  ^  =  57°  13',  B  =  67°  53'. 

3.  Show  that  the  double  area  of  any  quadrilateral  is  the  product  of 
its  two  diagonals  into  the  sine  of  the  angle  between  them. 

4.  Show  that  the  perimeter  of  any  triangle  is  the  diameter  of  the 
circumscribing  circle  into  the  sum  of  the  sines  of  its  angles,  and  the  area 
of  any  triangle  is  the  product  of  the  radii  of  the  inscribed  and  circum- 
scribed circles  into  the  sum  of  the  sines  of  its  angles. 

5.  Show  that  if  triangles  of  equal  area  are  circumscribed  about  the 
same  circle,  their  perimeters  are  the  same.  What  is  the  sum  of  the  sines 
of  their  angles  ? 

6.  The  diagonals  of  a  quadrilateral  are  150,  131,  and  their  angle 
45°  15' ;  what  is  its  area? 


§91] 


THE   SINE   FAMILY. 


171 


7.  The  diagonals  of  a  parallelogram  are  123,  321 ;  find  its  sides,  angles, 
and  area,  when  the  inclination  of  the  diagonals  is  45°  25'. 

8.  The  sides  of  a  parallelogram  are  150,  130,  and  one  of  its  angles  is 
37°  15' ;  find  its  area. 

9.  Construct  examples  similar  to  Ex.  2  for  one-figured  data;    two- 
figured  ;  four-figured ;  five-figured.     Solve.     Test. 


§91.    To  Determine  the  Sine  of  Any  Angle  of  a  Triangle,  and 
the  Area  of  the  Triangle,  when  the  Three  Sides  are  Given. 

Let  ABC  he  the  given  triangle,  with  CD  perpendicular  to 
AB,  and  lengths  as  in  the  diagram. 


P*=b2-X>,                                      (1)                                  ( 

■» 

p  =  a2  -  (e  -  xy.               (2)                    / 

...  a2  -0  -  x)2  =h2-x2.        (3)             y     P 

\a 

.\  a2  -  <?  +  2  ex  =  b\                 (4^         / 

A         x        D    c-x  B 

.    x  _  ?)2  +  ^  -  a2                         f$\                    Fio.  7fi. 
*j  c 

■  by  (1), 

f=(b  +  x)(b-x) 

-K+,'r*)(»-,,+,'.-') 

2be  +  b2  +  c2-a?    2bc-b2-<?  +  a2 

2c                            2c 

_(h  +  cy-a2    a2-(b-c)2 

2c                     2c 

(b  +  c  +  a)(b  +  c  —  a)(a  —  b  +  c)(a  +  b  —  c) 

(6) 

4c2 

Let                          a  +  b-+c  =  2s.                                   .    (7) 

Then                   _a+5  +  (7==2(s-a),                             (8) 

a-b  +  c=2(s-V),                             (9) 

a +  &-<?  =  2(s-<?). 

(10) 

172 

Now, 

By  symmetry, 
By  §  90, 


PLANE   TRIGONOMETRY. 


sin  A  =  |-. 
o 


2 


sin  A  =  £-  vs(s  -a)(s-  b) (s-c). 

DC 


sin  J5  =  —  V«(s  -  a)  (s  -  b)  (s-c), 
ac 


sin  C  =  -4-  ^s(s  -  a)(s-  b)(s  -  c). 
ab 


[§91 

(11) 
(12) 

(13) 


A  =s  area  of  triangle  =  — '— ,  etc. 

A 

A=  y/s(s-a)(s-b)(s-c)> 


(10)  may  be  used  to  find  the  area  of  a  triangle  when  the 
three  sides  are  given.  (7),  (8),  (9)  may  be  used  to  find 
the  sines  of  the  angles.  The  angles  themselves  are  not 
uniquely  determined,  since  two  angles  less  than  180°  have 
the  same  sine.  For  this  reason  any  angle  is  found  from  the 
sine  of  its  half  rather  than  from  the  foregoing  expressions, 
since  the  half  angle  is  acute. 


§  92.    The  Sine  of  the  Half  Angle  of  a  Triangle  in  Terms  of 

the  Sides. 

Inscribe  in  the  given  triangle 
a  circle  (Fig.  77). 

AD  =  AF,  BD  =  BE,   OF=OE. 
.-.  2AD  +  2BO 

=  perimeter  of  triangle.    J, 
.-.  AD  =  s-a.  (1) 

The  three  triangles  A  OB,  BOO,  COA  make  up  the  tri- 
angle ABO, 


r(a  +  b  +  <?)  =  2  A  =  2  Vs(s  -«)(«-  6)(s  -  <?)■ 


Now, 


...  r-  J(g-a)(*-5)<>-g) 

*  8 

.   A    DO      r 

2      AO     AO' 


(2) 
(3) 


93]  THE   SINE   FAMILY.  173 

But  ZO2  =  r2  +  A& 

(s  —  a)(s  —  6)(s  —  (?) 


+  (*-«)*    by  (2),  (1) 


—  a 


{(t  _*)(•- 0>+ •(•--•)! 


C!_^{(a_(6.c)Xa  +  (ft_c))  +  (ft  +  c  +  aXft+c_a)j 


•  5c,  when  multiplied  out. 


v^* 


)(«-ft)(«-<0 


.  sin#=-^  = 8       =J<« -»)(«-«).     (4) 


sin  |  =  V(8~a^8~C)>    (5),  by  symmetry. 


sin  %=^{8~a)^~b\   (6),  by  symmetry. 


§  93.    Calculation  of  the  Half  Angles  of  a  Triangle  when  the 
Three  Sides  are  Given. 

It  is  best,  when  using  logs,  to  set  the  sines  in  the  forms 


n\„  A  -  ~  K*  -  a)o  -  &x*  - g)      a 


:~^- J(« -«)(«- 6)(«-<0         b 


sin 


=Vi£^£ 


abc  s  —  b 


Zinc-J(°-  *)(»  -  b-)C  -  o)  .  _J_. 
2       *  afo  « —  c 

The  scheme  of  solution  will  then  be,  the  numerals  giving 
the  order  of  filling  in,  all  logs  being  looked  up  before  manipu- 
lation begins : 


174  PLANE  TRIGONOMETRY.  [§93 

(1) 

b=  (2) 

c=  (3) 

2«=  (4) 

•  -  (5) 

a  —  a  =  (6) 

.-J  =  (7) 

s-e=  (8)     CAecft.-  (6)  +  (7)  +  (8)  =  (5) 

log  («-«)=  (9) 

log(»-6)=  (10) 

log  (•-«)=  (11) 
log  (s  — a)(s  — 6)(s  — c) 

(15),  by  adding  (9),  (10),  (11) 

\oga=  (12) 

log  6=  (13) 

log*=  (14) 

log  «6(?  =  (16),  by  adding  (12),  (13),  (14) 

logO-a)O-ft)<>-<0 

abc 

=  (17),  by  subtracting  (16)  from  (15) 

log— —  =  (18),  by  subtracting  (9)  from  (12) 

8  —  a 

log  -A-  =  (19),  by  subtracting  (10)  from  (13) 

8  —  0 

log— £_=  (20),  by  subtracting  (11)  from  (14) 

»  —  <j 

2  log  sin  ^  =         (21),  by  adding  (17),  (18) 


2  log  sin  |  =         (22),  by  adding  (17),  (19) 
2  log  sin  £  =         (23),  by  adding  (17),  (20) 
log  sin  4  =  (24),  by  taking  half  of  (21) 


(27) 
(28) 
(29) 


§  93]  THE   SINE  FAMILY.  175 

log  sin  ^  =         (25),  by  taking  half  of  (22) 

log  sin  %  =         (26),  by  taking  half  of  (23) 

A  = 

"   2 

B 

2 

2 

.-.  A  =         (30) 
B=         (31) 
(7=         (32) 
Test.     Sum  =180°,  (33),  approximately. 

EXERCISES. 

Determine,  to  the  appropriate  reading,  the  angles  of  the  following  tri- 
angles, by  means  of  the  sines  of  the  half  angles.  Test  by  adding  the 
angles.  Their  sum  should  be  approximately  180°,  for  data  of  three  or 
more  figures.  When  four-place  tables  are  used  on  four-figured  data,  the 
logarithm  of  the  sine  for  the  half  angle  may  be  in  error  by  almost  1  in 
the  fourth  place.  (See  §  24,  3.)  By  looking  at  a  four-place  table, 
it  will  be  observed  that  for  angles  between  6°  and  10°,  the  difference 
for  1'  is  about  9,  and  that  when  the  angle  is  about  45°,  the  difference 
for  1'  is  about  1.  The  effect  on  the  angle,  then,  of  an  error  of  1  in 
the  fourth  place  of  logs  varies  with  the  size  of  the  angle.  By  a  glance 
at  the  table  for  angles  in  the  neighborhood  of  the  angles  calculated, 
noting  what  the  angle  difference  is  there  for  an  error  of  1  in  the  fourth 
place  of  logs,  the  student  can  at  once  see  in  any  special  case  whether  the 
sum  of  the  calculated  angles  is  sufficiently  near  180°.  For  example,  if 
the  triangle  is  approximately  equilateral,  each  half  angle  is  about  30°. 
In  that  part  of  the  table,  1  in  the  fourth  place  of  logs  corresponds  to 
about  half  a  minute  in  the  angle.  So  that  each  angle  of  the  triangle 
may  be  in  error  by  about  1'  on  account  of  the  tables.  All  the  calcula- 
tions may  thus  be  correct,  and  the  sum  of  the  angles  differ  from  180° 
by  as  much  as  3'.  Similar  considerations  hold  for  any  place  table.  The 
angles  may  sum  to  180°  and  each  angle  be  wrong  by  balancing  errors. 
A  log  test  should  be  used. 


176  PLANE   TRIGONOMETRY.  [§94 

1.  The  triangle  has  the  sides  23,  25,  30 ;  4,  5,  7. 

2.  The  triangle  has  the  sides  234,  240,  251. 

3.  The  triangle  has  the  sides  34.5,  35.6,  289. 

4.  The  triangle  has  the  sides  34.51,  22.43,  16.85. 

5.  The  triangle  has  the  sides  345.3,  185.3,  298.7. 

6.  The  teacher  may  assign  data  suitable  for  a  five-place  table. 

7.  For  one-figured  data  on  lines,  how  close  should  the  test  A  +  B  + 
C  =  180°  hold? 

8.  How  close  for  two-figured  data? 

9.  How  close  for  three-figured  data? 

10.  How  close  for  four-figured  data  ? 

11.  How  close  for  five-figured  data  ? 

12.  How  close  for  six-figured  data  ? 

13.  How  close  for  seven-figured  data  ? 

§  94.     Use  of  the  Sine  in  constructing  Given  Angles. 

It  is  frequently  necessary,  in  solving  problems  in  mechan- 
ics by  graphical  methods,  to  lay  out  angles  of  given  size  more 
accurately  than  can  be  done  by  means  of  a  protractor  (sub- 
ject to  errors  of  warping,  centring,  alligning,  etc.).  For 
angles  not  very  small  the  sine  may  be  used  advantageously. 

In  Fig.  78  the  chord  of  the  arc  AMB  is  twice  the  radius 
times  the  sine  of  the  half  of  the  angle  A  OB, 

AB=2  AN=  2  OA  •  sin  AON. 

Thus,  if  OA  is  the  unit  of  the  scale 

of  the  diagram,  twice  the  sine  of  the 

half  angle  taken  from  the  tables  will 

be  the  chord  AB,  to  the  same  scale. 

Fig.  78.  Thus,  to  lay  out  a  given  angle  at 

a  given  point  on  a  given  line,  draw 

about  the  given  point  as  centre,  a  circle  whose  radius  is  the 

scale  unit.     From  the  point  where  this  arc  cuts  the  given 

line  as  a  centre  and  with  twice  the  sine  of  half  the  given 


§95]  THE   SINE   FAMILY.  177 

angle,  on  the  same  scale  as  radius,  draw  an  arc  cutting  the 
given  arc.  Join  this  point  of  intersection  of  arcs  and  the 
given  vertex  point. 

EXERCISES. 

The  teacher  may  make  selections. 

§  95.     The  Reciprocal  Sine,  or  Cosecant. 

The  reciprocal  of  the  sine  is  - —     This  is  called  the  co- 
sine 

secant.     The  cosecant  of  x  is  written  cosec  x,  or  esc  x. 

The  practical  importance  of  the  cosecant  is  small,  for  mul- 
tiplying by  it  is  the  same  as  dividing  by  the  sine,  and  vice 
versa.  Also  log  cosec  x  =  —  log  sin  x.  For  these  reasons  it 
is  neither  customary  to  give  the  cosecant  in  tables,  nor  to 
use  it  in  calculations.     "  Cambria  Steel "  gives  cosecants. 

One  must  not  confuse  the  reciprocal  sine  with  the  inverse 
sine  (§  73)  ;  that  is,  (sin  x)~l  with  sin"1  x. 

The  sign  of  the  cosecant  is  evidently  the  same  as  that  of 
the  sine. 

The  range  of  values  of  the  cosecant  is  very  different  from 
that  of  the  sine.  As  the  angle  runs  from  0°  to  360°,  the  sine, 
it  will  be  remembered,  runs  through  the  continuum  from  0 
to  1  to  0  to  —1  to  0.  ,At  the  same  time  the  cosecant  runs 
through  the  reciprocal  set  of  values:  from  plus  infinity  to 
unity,  then  back  to  plus  infinity;  then  suddenly  changes 
from  plus  infinity  to  minus  infinity ;  then  goes  on  continu- 
ously to  minus  unity;  then  back  to  minus  infinity,  as  the 
angle  approaches  360°.  As  the  angle  passes  360°,  there  is  a 
sudden  change  in  value  of  the  cosecant  from  minus  infinity 
to  plus  infinity.  Then  there  is  a  repetition  of  what  has 
just  happened  from  0°  to  360°,  etc. 

The  cosecant  is  periodic  with  the  same  period  as  the  sine ; 
but  while  the  sine  is  a  continuous  function  of  the  angle,  the 
cosecant  is  discontinuous,  there  being  jumps  from  plus  infinity 
to  minus  infinity  when  the  angle  passes  through  odd  multi- 


178 


PLANE   TRIGONOMETRY. 


[§96 


pies  of  180°,  and  from  minus  infinity  to  plus  infinity  when 
the  angle  passes  through  any  even  multiple  of  180°,  counter- 
clockwise motion  being  assumed. 


§  96.    Line  Pictures  of  the  Cosecant  in  Each  Quadrant. 


B 

S 

\g£>^ 

^J^if 

\             ° 

MJ 

N 

sf 

B 

N      \M 

0              J 

Fig.  79. 


Fig.  80. 


Fig.  81. 


Fig.  82. 


In  the  unit-circle  (Figs.  79,  80,  81,  82)  MP  is  the  sine, 
and,  consequently,  OS  the  cosecant. 


^         MP     ITS        MP 

For      op=WorT" 


os 


-.  OS 


MP 


Since  MP  is  the  sine,  OS  is  the  cosecant. 

The  cosecant  is  obtained  by  drawing  the  tangent  at  the 
90°  point  of  the  unit  circle,  and  prolonging  it  to  meet  the 
terminal  of  the  angle.  In  the  first  and  second  quadrants 
the  direct  terminal  is  met,  and  the  cosecant  is  positive.  In 
the  third  and  fourth  quadrants  it  is  the  opposite  terminal 
which  is  met,  and  the  cosecant  is  negative. 


97] 


THE   SINE   FAMILY. 


179 


§  97.   Line  Picture  of  the  Cosecants  of  All  Angles. 


Fig.  83. 

For  the  position,  OA,  of  the  terminal  the  cosecant  is  positive 
infinity  (+  oo).  It  diminishes  as  the  terminal  turns,  counter- 
clockwise, taking  the  values  (where  OB  =  1) 

+  oo  -  0S±  ..  0S3  -  0S2  .•  0S1  -  OB. 

Then  the  values 

OB  -.  0S_X .-  0S_2  -.  0S_3  -  +  oo, 

as  the  terminal  goes  from  90°  to  180°. 

As  the  terminal  passes  through  180°,  there  is  the  discon- 
tinuity from  +  oo  to  —  oo,  since  the  cosecant  passes  to  the 
opposite  terminal,  and  the  values  indicated  above,  with  signs 
changed,  are  passed  over,  as  the  terminal  moves  from  180°  to 
360°. 

As  the  terminal  passes  360°  there  is  again  a  discontinuity, 
from  —  oo  to  4-  oo,  since  the  cosecant  passes  from  the  opposite 
terminal  to  the  direct  terminal. 

As  the  terminal  continues  to  turn  there  is  a  repetition  of 
what  has  just  taken  place. 


EXAMPLES. 
1.   Solve  the  equation  sin  x  +  cosec  x  =  f . 

Hint  :  Let  cosec  x  = Solve  the  resulting  quadratic  for  sin  x. 

sin  a; 
Recall  the  corresponding  angles. 


180  PLANE   TRIGONOMETRY.  [§98 

2.  Can  there  be  an  angle  whose  cosecant  is  £?  What  is  the  least 
value  of  the  cosecant  ?  When  the  sine  increases,  does  the  cosecant  increase 
or  decrease  ? 

3.  Construct  an  angle  whose  cosecant  is  f ;   —  f . 

4.  From  the  corresponding  matter  in  connection  with  sines,  read 
the  following  expressions: 

cosec-1  f ;     cosec-1  (  —  f )  ;     2  cosec-1 1 ;     3  cosec-1  (  —  f ) . 

Construct  the  terminals  for  these  expressions.     How  many  terminals 
for  any  given  value  of  the  cosecant  ? 

5.  Give  the  general  value  of  all  angles  which  have  the  same  cosecant 
as  any  given  angle  A,  a. 

6.  What  is  the  relation  of  the  terminals  of  angles  which  have  the 
same  cosecant  ?    Opposite  cosecants  ? 

7.  Determine  and  tabulate  the  cosecants  corresponding  to  the  sixteen 
terminals  of  §  75. 

8.  Solve  the  equation  sin  x  +  cosec  a;  =  —  §  (general  value). 

3  V2 

9.  Solve  the  equation  sin  x  +  cosec  x  =  ± (general  value). 

10.  Solve  with  tables  the  equation  sin  x  +  3  cosec  x  =  3.1432. 

11.  Solve  the  equation  4  sin  0  =  cosec  6  (general  value) . 

12.  Solve  the  equation  4  sin  0  =  3  cosec  6  (general  value). 

13.  Solve  the  equation  sin  0  +  3  cosec  6  =  \  (general  value). 

14.  Solve  the  equation  3  sin  0—8  cosec  0  = ■ (general  value). 

15.  Solve  the  equation  sin  a;  +  cosec  x  =  2.5618. 


§  98.     The  Graph  of  a  Function  in  Rectangular  Coordinates. 

Suppose  y  is  a  single-valued  function  of  a?,  and  that  for 
any  arbitrary  value,  a,  of  #,  y  has  been  calculated,  giving  b. 
This  pair  of  corresponding  values  may  be  taken  as  the  co- 
ordinates of  a  point  in  a  plane,  the  x  as  abscissa,  and  the  y 
as  ordinate.  Let  this  point  be  constructed.  Suppose  it  is 
A  in  Fig.  84,  with  the  abscissa  0M=  a  and  the  ordinate 
MA  =  b.  Suppose  a  second  point,  B,  obtained  in  the  same 
way,  with  x  —  ON '=  c,  y  =  NB  =  d.  If  we  imagine  now  that 
x  takes  all  values  of  the  continuum  from  x  =  a  to  x  =  c,  and 


98] 


THE   SINE   FAMILY. 


181 


~r  ~r 

x 

>                      ,--«,. 

i       -/     *£ 

A-.              Z 

s       ^ 

I 

-  0                 if                                     M 

Fig.  84. 


that  the  corresponding  values  of  y  are  calculated  and  the 
corresponding  points  plotted,  as  were  A,  B,  we  shall  have 
on  the  paper  a  set  of 
points  which  viewed  as 
an  assemblage  will  con- 
stitute a  line  from  A  to 
B,  straight  or  zigzag  or 
crinkly  or  curved,  or, 
as  commonly  called,  a 
curve  (this  term  includ- 
ing also  straight  lines  as 
a  special  case).     Such  a 

curve  is  called  the  graph  of  the  function,  y,  from  A  to  B. 
In  practice  we  cannot  construct  an  infinite  number  of  points. 
A  sufficiently  large  number  of  points  to  give  us  a  more  or 
less  close  approximation  to  the  true  curve,  depending  on  the 
use  that  is  to  be  made  of  the  result,  are  plotted  and  joined 
by  a  smooth  curve.  (In  practice  a  so-called  "  French  Curve  " 
is  used.)  The  graph  shows  to  the  eye  the  relation  between 
the  function,  y,  and  the  variable,  x,  both  as  to  corresponding 
magnitudes  and  as  to  rate  of  change  of  the  function  as  the 
variable  changes.  The  graph  of  a  function  is  frequently 
called  the  locus  of  the  equation  connecting  x,  y ;  that  is,  it  is 
the  places  of  all  the  points  whose  coordinates  give  pairs  of 
values  of  x,  y,  which  satisfy  the  equation.  A  graph  is  also 
called  a  locus  of  a  point,  moving  in  a  specified  manner. 

In  the  case  of  a  two-valued  function,  each  value  of  x  will 
give  two  values  of  y.  So  two  points  will  have  the  same 
abscissa.  The  corresponding  curve,  or  graph,  will  have 
two  branches.  Similarly,  an  n-valued  function  will  have  a 
graph  of  n  branches. 

A  graph  can  also  be  used  to  show  pictorially  the  connec- 
tion between  quantities  whose  relation  to  each  other  is  not 
expressible  in  the  form  of  an  equation.  For  example,  we 
may  take,  from  the  census  reports,  the  population  of  a  city 
for  every  ten  years  for  a  number  of  years.  Laying  out  the 
years,  to  any  scale,  as  abscissas,  and  the  population  as  corre- 


182 


PLANE   TRIGONOMETRY. 


[§98 


sponding  ordinates,  to  any  convenient  scale  (500  inhabitants 
to  an  inch,  or  any  other  convenient  number),  then  connect- 
ing the  plotted  points  by  a  smooth  curve,  we  have  the  popu- 
lation graph.  The  ordinate  for  any  selected  year  will  give 
the  population,  approximately,  for  that  year.  The  graph 
will  thus  enable  one  to  calculate  the  population  fairly  well 
for  years  other  than  the  census  years. 

Such  graphs  are  used  extensively  in  engineering  work, 
also  in  economic  reports  for  the  exhibition  of  statistics. 
They  tell  the  whole  story  pictorially,  and  enlighten  where 
the  same  report  given  in  figures  would  confuse  or  make  no 
impression  at  all. 


ILLUSTRATIVE   EXAMPLES. 
1.  y  =  x. 

This  is  evidently  satisfied  by  any  point  on  a  straight  line  passing 
through  the  origin  and  bisecting  the  first  and  third  quadrants.     This 

line  is  the  graph. 

2.  y  =  -x. 

This  is  satisfied  by  any  point  on 
the  bisector  of  the  second  and  fourth 
quadrants. 

3.  y  =  x  +  1. 
This  is  evidently  satisfied  by  any 

point  on  a  line  passing  through  the 
point  (0,  1)  on  the  ordinate  axis  and 
making  an  angle  of  45°  with  abscissa 
axis. 

4.  y  s  x  —  1. 

This  is  a  straight  line  passing 
through  (0,  —  1)  on  the  ordinate  axis  and  making  45°  with  the  abscissa 
axis. 

5.  x  =  ±2. 

This  is  a  pair  of  straight  lines  parallel  to  the  ordinate  axis  and  at 
distances  2  to  the  right  and  left  of  the  origin. 

6.  y  =  ±2. 

This  is  a  pair  of  lines  horizontal  and  at  distances  2  above  and  below 
the  abscissa  axis. 


f 

/> 

X 

1 

f, 

{) 

Fig.  85. 


98] 


THE   SINE  FAMILY. 


183 


7.   x2  +  y*  =  25. 

This  is  satisfied  by  any  point  on  a  circle  of  radius  5,  with  its  centre 
at  the  origin.  This  is  an  example  of  a  two-valued  function,  since  for 
any  x  there  are  two  y's, 


V- 


y  =±V2o-x2. 
Thus,  there  are  two  points  equidistant 
from  the  abscissa  axis  which  satisfy 
the  given  equation,  both  points  having 
the  same  abscissa,  or  x.  Similarly  for 
x  in  terms  of  y. 

8.  The  preceding  examples  are 
those  of  functions  whose  graphs  can 
be  seen  readily  without  calculation. 
To  illustrate  the  calculation  process, 
we  may  take 

y2  =  4  x.  Fig.  86. 

Giving  to  x  the  values  following,  one  can  calculate  the  corresponding  y's, 
or  get  them  from  a  table  of  squares : 


\ 


X 

±y 

0 

0 

.1 

.6 

.2 

.9 

.3 

1.1 

.4 

1.3 

.5 

1.4 

.6 

1.5 

.7 

1.7 

.8 

1.8 

.9 

1.9 

X 

±y 

1.0 

2.0 

1.1 

2.1 

1.2 

2.2 

1.3 

2.3 

1.4 

2.4 

1.5 

2.5 

1.6 

2.5 

1.7 

2.6 

1.8 

2.7 

1.9 

2.8 

X 

±y 

2.0 

2.8 

2.2 

3.0 

2.4 

3.1 

2.6 

3.2 

2.8 

3.3 

3.0 

3.5 

3.2 

3.6 

3.4 

3.7 

3.6 

3.8 

3.8 

3.9 

X 

±1 

4.3 

4.1 

4.6 

4.3 

4.9 

4.4 

5.2 

4.6 

5.5 

4.7 

5.8 

4.8 

6.1 

4.9 

- 

These  values  plotted  to  scale  (1  inch  =  1,  say  ),  taking  y  to  the  nearest  10th, 
give  the  curve.    The  student  may  plot  this  curve.    Use  coordinate  paper. 

9.  Construct  the  graphs  of  the  following : 
#2  =  -4x,   x2  =  4y,   x2  =  -iy,   y  =  2x+3,   y  =  3x-4,   y  =  -2x  +  3, 
y=-2x-l,   3x  +  2y  =  6,   3x-2y  =  6,     -3x+2y  =  6,     -3x-2y  =  6. 

10.  Look  up  in  the  census  reports  the  population  of  the  three  most 
important  cities  in  your  state,  and  construct  corresponding  graphs  to  the 
same  scale  and  compare.  Calculate  from  the  graphs  the  population  for 
some  year  not  a  census  year. 

11.  On  rectangular  coordinate  paper  make  graphs  of  y  =  log10a:  and 
V  =  1°&«  x->  to  the  same  scale.  What  is  the  relation  between  correspond- 
ing #'s  for  the  same  #?     (See  §  5.) 


184 


PLANE   TRIGONOMETRY. 


[§99  a 


§  99.    Graph  of  the  Sine,  or  Locus  of  the  Equation/  =  sin  Jr. 

(a)  Imagine  the  unit  circle  made  of  wire,  with  each  ordi- 
nate a  wire  hinged  at  its  extremity  to  the  unit  circle.  Clip 
the   circle  wire  at  a,  and  stretch  it  out  straight,  with  the 

positive  ordinates  pointing  up 
and  the  negative  ordinates 
down.  The  ends  of  the  wire 
ordinates  give  the  graph  of  the 
sine  for  all  values  of  x  from 
zero   to   2-7T  radians. 

(5)  An  easy  practical  process 
for  making  the  graph  of  the  sine. 
Draw  a  circle  of  any  radius  on 
a  piece  of  pasteboard.  Draw 
on  it  a  large  number  of  ordi- 
nates. Cut  the  circle  out  with 
a  sharp  knife,  so  as  to  give  a  smooth  edge.  Roll  the 
circle  in  a  vertical  position  on  a  straight  line  of  a  horizontal 
sheet  of  paper.  Dot  the  line  with  a  pencil  at  the  places 
where  the  ends  of  ordinates  touch  it,  laying  out  at  such  dots 
the  corresponding  ordinates,  in  length  and  sign.  Join  the 
ends  by  a  smooth  curve.  This  is  the  sine  graph  corre- 
sponding to  the  radius  of  the  selected  circle  as  unit. 


T-£-\ 

£"g  c 

-"■^l 

3^^ 

■^g 

•  / 

V*. 

~3Z 

Sb 

J 

\ 

r 

i 

1 

c 

[_ 

] 

k 

d 

t 

i 

5 

t 

v 

\ 

7 

s 

z 

s 

/ 

S* 

^ 

^.~*> 

x: 

Fig.  87. 


Fig.  88.  —  Sine  graph,  x  =  0  to  x  =  2  ir. 


(<?)  Continuous  graph  of  the  sine.  Imagine  the  preceding 
curve  (Fig.  88)  repeated  indefinitely  to  the  right  and  left ; 
we  have  the  sine  graph  for  all  values  of  x  from  minus  infinity 
to  plus  infinity.     (See  another  method  in  Ex.  8,  page  191.) 


§99/]  THE   SINE   FAMILY.  185 

After  the  curve  from  a  to  the  end  of  the  central  ordinate 
at  /  has  been  plotted,  a  pasteboard  cutting  of  this  may  be 
made  and  used  to  continue  the  curve  indefinitely  forwards 
and  backwards. 


The  curve  of  Fig.  89  shows  the  periodicity  of  the  sine,  by 
the  equality  of  ordinates  separated  by  distances  which  are 
multiples  of  2  w.     (See  §  72.) 

(d)  The  units  in  the  graph  of  the  sine.  The  angle  must  be 
expressed  in  circular  measure,  for  the  sines  (the  ordinates)  are 
expressed  as  ratios  in  terms  of  the  modulus,  or  radius.  Con- 
sequently the  horizontal  distances  (the  angles)  of  the  graph 
must  be  expressed  in  terms  of  the  same  unit  (the  radius), 
that  the  graph  may  give  a  correct  impression  of  the  relation 
of  the  two  connected  quantities,  the  sine  and  angle. 

Thus  57°.3  is  the  unit  for  horizontal  distances  when  using 
degrees.  The  same  length  which  represents  57°.3  will  rep- 
resent the  sine  of  90°,  while  the  horizontal  distance  for  90° 

90 
will  be  r=-<j.    Thus  the  height  of  the  curve  will  be  less  than 
57.  o 

half  the  distance  between  the  points  where  it  crosses  the 
horizontal  line. 

(e)  Using  the  table  of  natural  sines  to  construct  the  sine 
curve.  Take  57°  18'  as  the  unit  (an  inch,  say,  on  the  scale). 
Tabulate  the  sines  of  all  eighths,  or  tenths,  of  57°  18',  to  run 
from  0°  to  90°.  Plot  corresponding  values.  Join  plotted 
points  by  a  smooth  curve.  Make  pasteboard  cutting  of  this 
and  use  it  to  continue  the  curve. 

(/)  The  sine  curve  as  a  wave.  The  sine  curve  is  of  great 
importance  in  mathematical  physics.  Many  sorts  of  physi- 
cal phenomena,  periodic  in  their  character,  can  be  repre- 
sented by  a  sine  curve,  or  by  two  or  more  sine  curves,  in  the 
same  plane,  or  in  different  planes,  combined  with  each  other. 

If  we  make  graphs  of  ?/  =  sm#  (1)  and  y  =  sin  (x  +  a)  (2) 


186  PLANE  TRIGONOMETRY.  [§99/ 

and  compare  them,  the  graph  of  (2)  will  be  that  of  (1)  slid 
along  the  horizontal  axis  a  distance  a,  forward  or  back- 
ward, according  to  the  sign  of  a.  The  ordinate  of  (2)  at 
any  point  is  that  of  (1)  at  a  distance  a  to  the  right  or  left, 
according  as  a  is  minus  or  plus.  Two  such  sine  curves 
are  said  to  be  in  different  phase,  and  a  is  called  the  phase- 
differ  enee.  The  effect  of  superposing  two  curves  is  obtained 
by  plotting  them  in  connection  with  the  same  pair  of  co- 
ordinate axes,  and  then  making  a  new  curve  whose  ordinate 
at  any  point  is  the  algebraic  sum  of  those  of  the  two  curves. 
Similarly  for  any  number  of  curves. 

Two  equal  sine  curves  are  said  to  be  in  opposite  phase 
when  their  ordinates  at  the  same  point  of  the  abscissa  axis 
are  equal  and  opposite.  If  two  such  curves  are  superposed, 
the  resulting  curve  is  the  horizontal  axis.  This  has  its 
counterpart  in  two  waves  of  sound  producing  silence.  If 
two  sine  curves  of  slightly  different  phase  are  superposed, 
the  hills  and  valleys  of  the  resulting  curve  will  be  more  pro- 
nounced than  those  of  the  given  curves.  This  has  its  physi- 
cal representative  in  the  familiar  phenomenon  of  "beats," 
when  two  musical  notes  are  "slightly  out  of  tune." 

Many  different  waves  may  be  superposed,  —  waves  dif- 
ferent in  phase  and  in  size  and  period  and  plane.  Such 
are  ocean  waves.  The  topic  is  an  extensive  one,  and  is  dis- 
cussed at  length  in  mathematical  treatises  on  sound,  light, 
electricity.  Here  we  have  given  merely  a  hint.  It  plays 
an  important  practical  role  in  the  discussion  of  Alternating 
Currents,  as  in  Steinmetz's  book  on  this  subject. 

§  100.     Harmonic  Motion. 

Imagine  a  terminal  line  turning  with  uniform  angular 
velocity  of  a  units  per  unit  of  time  (a  second,  generally). 
At  the  end  of  t  time  units,  counting  time  from  the  moment 
when  the  terminal  coincides  with  the  initial  line,  the  angle 
A  OP  in  Fig.  90  is  at     Let  MP  be  the  ordinate  from  P. 

.\  MP  =  r  •  sin  («Q.  (1) 


§  100]  THE   SINE   FAMILY.  187 

If  instead  of  counting  time  from  the  moment  when  OP 
coincides  with  OA,  it  is  counted  from  the  position  ON  (/3), 

MP  =  r  •  sin  (at  +  ff).  (2) 

Draw  now  PP1  perpendicular  to  the  vertical  axis, 
called  the  projection  of  P  on  this  axis. 
As  P  moves  uniformly  around  the  circle, 
Px  moves  back  and  forth  along  the  ver- 
tical axis.  The  motion  of  Px  is  said  to 
be  harmonic,  being  similar  to  that  of  a 
point  in  a  vibrating  string,  or  to  that 
of  the  wood  particles  in  a  piano  sound- 
ing board. 

The  radius  of  the  circle,  or  the  length  FlG#  90' 

r  in  (1),  (2),  is  called  the  amplitude  of  motion,  being  the 
half  swing  of  the  point  Pv 

If  graphs  were  made  for  (1),  (2),  r  would  be  the  wave 
height. 

If  one  has  before  him  a  graph  of  (1)  or  (2),  and  imagines, 
as  he  gazes  at  the  graph  on  its  page,  that  the  graph  travels 
across  the  page  with  the  flow  of  time  and  at  the  rate  of 
a  units  per  second,  he  has  a  picture  of  a  simple  ocean  wave. 
Here  a  would  be  called  the  wave  velocity  instead  of  the 
angular  velocity.  If  instead  of  watching  the  whole  graph 
move,  the  observer  fixes  his  eye  upon  a  single  point  of  the 
horizontal  axis  and  watches  the  changing  height  of  the  or- 
dinates  at  that  point,  as  the  moving  graphs  pass  it,  he  has 
a  picture  of  harmonic  motion  at  a  point.  It  is  approxi- 
mately the  motion  of  the  straight  part  of  the  piston  rod  of  a 
stationary  engine  as  the  driving  wheel  turns  uniformly. 

/3  of  (2),  without  reference  to  (1),  is  called  the  phase  at 
zero-time.  It  is  the  position  of  the  terminal  at  zero-time, 
in  considering  harmonic  motion.  In  the  sine  graph,  it  is 
the  angle  whose  sine  is  the  ratio  of  the  wave  height  on  the 

vertical  axis  at  the  origin  to  the  total  wave  height. 

o  _ 

In  (1),  (2), is  called  the  periodic  time;  for  if  in  (1), 


188  PLANE  TRIGONOMETRY.  [§100 

2  7T 

(2),  t  is  changed  by  ,  then  calling  the  time  at  the  first 

instant  tx  and  that  at  the  second  tv 

2tt 

The  ordinate  for  tx  is  r  •  sin  (a^  +  /3). 
The  ordinate  for  t2  is  r  •  sin  (ctf2  +  /3)  ; 

or,  r  •  sinO/^  H J  +  0)  ; 

or,  r  •  sin  (a^  +  y8  +  2  7r)  ; 

or,  r  •  sin  (atx  +  /3) 

.*.  the  ordinates  are  equal. 

This  is  also  evident  by  considering  the  uniform  turn  of  OP. 
The  time  of  a  complete  turn,  if  the  rate  is  a  units  per  second, 

2  7T 

is — ,  a  being  in  radian  measure.     After  a  complete  turn  of 

OP  from  any  position,  Px  is  again  in  the  same  state  of  mo- 
tion as  when  OP  formerly  occupied  this  position. 

This  is  but  a  touch  on  a  topic  of  great  practical  importance 
in  the  discussion  of  alternating  electrical  currents,  light,  etc. 

§  101.     Graph  of  the  Cosecant. 

The  graph  of  the  cosecant  is  readily  constructed  from  that 
of  the  sine,  since  sin  times  cosec  =  1.  In  Fig.  91,  represent- 
ing a  right-angled  triangle  in  a  circle, 
with  CM  perpendicular  to  AB,  AC2  = 
AB  •  AM,  from  the  similarity  of  the  tri- 
angles AMQ,  ABO. 

If  then  at  any  point  M,  Fig.  92,  of  the 
sine  graph,  we  draw  the  line  MO  perpen- 
dicular to  the  ordinate  AM,  and  with  A 
as  a  centre  and  the  height  of  the  sine 
curve  as  radius,  draw  an  arc,  cutting  MO  in  (7,  OB  drawn 


102]  THE  SINE  FAMILY. 

B 


189 


Fig.  92. 


at  right  angles  to  AC  will  cut  AM  prolonged  in  J9,  giving 
AB  as  the  length  of  the 
cosecant  at  the  point  A, 
and  B  as  a  point  on  the 
cosecant  graph. 

In  the  same  way 
any  number  of  points 
on  the  graph  can  be 
found. 

The  appearance  of  the 
cosecant  curve  as  re- 
lated to  the  sine  curve 
is  that  of  Fig.  93. 

Fig.  93. 

LABORATORY  EXERCISE. 

Construct  together  some  sine  curve  and  its  corresponding  cosecant 
curve.  Tilt  a  sine  curve,  to  give  an  anti-sine  curve.  Tilt  a  cosecant 
curve,  to  give  an  anti-cosecant  curve. 

§  102.     Graphs  in  Polar  Coordinates. 

When  an  equation  is  given  connecting  r,  0,  we  can,  by 
giving  one  of  them  special  values,  calculate  the  other. 
Plotting  a  number  of  corresponding  points,  not  too  far 
apart,  and  joining  them  by  a  smooth  curve,  we  have  the 
graph.  For  plotting  such  curves  Groat's  polar  coordinate 
paper  will  be  found  very  useful. 


+co      H-oo 

1              1 

1 

I                   1 

4-co      +  oo 

1             / 

\         / 

Sir 

2 

1             / 

w     T 

0  7T       7T     57T7I 

6      2    TT 

w 

27T     57T 

2 

Sl> 

\ 

/             I 
1                t 

\ 

-oo         —  CO 

1 

1    i 

—  OO            — OO 

1 1 

190  PLANE   TRIGONOMETRY.  [§  102 

What  is  r  =  a,  where  a  is  a  constant  ?  What  is  r*  —  5r  +  6 
—  0  ?  What  is  F(r)  =  0,  where  the  function  is  many  valued  ? 
What  is  0  =  0  ?  What  is  0  =  a,  where  a  is  a  constant  ?  What 
is  £2-50  +  6  =  0?  What  is  F(&)  =  0,  where  I7  is  many 
valued  ?, 

EXERCISES. 

1.  Use  Groat's  radian  polar  coordinate  paper  to  make  the  graph  of 
r  =  10  0,  r  =  20  0,  r  =  aO.  (Plan  :  give  values  to  0,  as  0.1  radian,  0.2 
radian,  etc.,  and  calculate  corresponding  values  of  r.  Then  plot  the 
corresponding  points  and  join  them  by  a  smooth  curve.  The  points  will 
be  for  the  first  curve:  (0,  0),  (1,  0.1),  (2,  0.2),  etc.  These  curves  are 
spirals  of  Archimedes.) 

2.  Use  Groat's  radian  polar  coordinate  paper  to  make  a  graph  of 

1  2  n 

r  =  -,  r  =  -,  r  =  --    (Hyperbolic  spirals.)     (Such  a  curve  with  r  about 

0  0  0  if 

an  inch,  or  inch  and  a  half,  when  0  =  ^  >  will  be  found  almost  as  useful, 
when  made  in  hard  wood,  for  drawing  purposes  as  a  "  French  Curve." 
Such  a  curve  was  found  most  useful  in  plotting  Professor  Orton's  Re- 
ports on  Ohio  Clays.)     Plot  r20  =  a.     (Lituus.) 

3.  Show  that  a  circle  is  the  graph  for  r  =  a  sin  0,  a  being  the  diameter 
(vertical),  0  being  measured  from  a  horizontal  tangent. 

4.  Use  Groat's  degree-measure  polar  coordinate  paper  to  plot  r  = 
10  sin  0  by  points,  taking  as  radial  unit  one  of  the  small  divisions  on  the 
paper.     See  if  your  curve  looks  like  a  circle. 

5.  Show  that  r  sin  0  =  a  is  a  straight  line  parallel  to  the  z-axis  at  a 
distance  a. 

6.  Use  Groat's  degree-measure  polar  coordinate  paper  to  make  a 
graph  for  r  sin  0  =  10. 

7.  Use  Groat's  degree-measure  polar  coordinate  paper  to  make  graphs 
of  r  sin  2  0  =  a  ;  r  =  a  sin  2  0;  r  sin  3  0  =  a ;  r  =  a  sin  3  0.  Give  a  some 
convenient  numerical  value.  See  if  the  number  of  loops  in  the  graph 
of  r  =  a  sin  nO  depends  upon  n  being  even  or  odd. 

8.  Draw  a  circle.  Take  a  horizontal  tangent  as  initial  line  and  its 
point  of  tangency  as  pole.  Prolong  each  radial  line  (vector)  from  this 
point  a  distance  equal  to  a  diameter  beyond  the  circle.  Connect  their 
ends  by  a  smooth  curve.  This  is  the  graph  then  of  r  =  a  (sin  0  +  1). 
It  has  the  shape  of  a  heart,  and  is  called  a  cardioid. 


§  103]  THE  SINE  FAMILY.  191 

9.   Use  Groat's  polar  coordinate  paper  (degree  measure)  to  plot 
r  =  a  (1  +  sin  0)  ;  r  =  a  (1  -  sin  0)  ;  r  (1  +  sin  0)  =  a  ;  r  (1  -  sin  0)  =  0, 
for  selected  values  of  a. 

10.  Draw  a  horizontal  line.  Select  a  point  above  it  at  a  distance  a 
as  pole.  Sketch  points  that  are  as  far  from  the  pole  as  from  the  line. 
The  curve  is  a  parabola.  Show  that  any  point  on  it  satisfies  the  equa- 
tion, r  (1  —  sin  6)  =  a.     What  is  r  (1  +  sin  6)  =  a? 

11.  Make  the  graphs  for  6  =  loge  r ;  6  =  log10  r ;  r  =  efl ;  r  =  10*. 

§  103.   The  Coversed  Sine. 

When  the  line  definition  was  in  use,  with  the  correspond- 
ing diagram  for  the  sine  to  the  radius  1,  the  balance  of  the 
radius  beyond  the  sine  was  given  the  name,  the  Coversed 
Sine.     And  still, 

coversin  A  =  1  —  sin  A. 

The  coversed  sine  is  used  in  engineering  field  books. 

EXERCISES. 

1.  Can  the  coversed  sine  be  negative  ? 

2.  "What  are  its  limits  of  value  ? 

3.  Construct  terminals  when  coversin  A  =  \ ;  coversin  A  =  f . 

4.  What  are  the  coversed  sines  for  angles  whose  terminals  border  the 
quadrants? 

5.  What  are  the  coversed  sines  for  angles  whose  terminals  are  the 
bisectors  and  trisectors  of  the  quadrants? 

6.  How  could  a  table  of  coversed  sines  be  made  for  angles  from  0°  to 
90°? 

7.  Make  the  graph  for  the  coversed  sine,  y  =  coversin  x. 

8.  Use  the  following  method  to  construct  a  sine-curve:  draw  a 
circle  of  one  inch  radius,  and  a  horizontal  line  through  its  centre,  A,  AO 
being  a  horizontal  radius  to  the  right.  Divide  the  circumference  into  9° 
spaces  from  O.  Lay  out  from  O,  right  and  left,  on  the  horizontal  line, 
to  the  extent  of  the  paper  used,  equal  distances,  each  0.16  inch  (9° 
in  radians,  inches).  Prolong  the  abscissas  of  the  extremities  of  the 
9°  divisions  to  meet  the  ordinates  at  the  ends  of  the  horizontal  divisions. 
Intersections  of  corresponding  lines  will  be  points  on  the  sine  curve. 
Use  a  similar  process  to  make  a  cosecant  curve. 


CHAPTER   VI. 


j 

> 

y 

/ 

< 

' 

\ 

) 

( 

) 

s 

1 

f 

THE    COSINE,     INVERSE     COSINE,     RECIPROCAL     COSINE 
(SECANT),  AND  VERSED  SINE  (1  -  COSINE)  OF  AN  ANGLE. 

§  104.  The  Cosine  of  an  Angle  is  the  ratio  of  the  abscissa  of 
any  point  on  the  terminal  of  the  angle  to  the  modulus  of  the  same 

point,  or  QM 

This  is  the  ratio-definition, 
or  Hassler's  definition.  The 
numerical  relation  of  an  angle 
in  circular  measure  to  its  cosine 
is  given  by  the  following  series, 
which  may  be  considered  also 
as  a  definition  of  the  cosine  : 

Similar  remarks  to  those  made  with  reference  to  the  defini- 
tions of  the  sine  hold  with  reference  to  those  of  the  cosine. 

As  with  the  sine,  the  cosine  has  a  threefold  use  in  cal- 
culations : 

(1)  When  the  abscissa  and  modulus  are  given,  their  ratio, 
expressed  as  a  decimal  fraction,  and  compared  with  a  table  of 
cosines,  will  indicate  the  corresponding  angle,  or  angles. 

(2)  When  the  modulus  and  angle  are  given,  the  cosine  of 
the  angle  is  the  number  which,  used  as  a  multiplier  with  the 
modulus,  will  give  the  abscissa. 

(3)  When  the  abscissa  and  angle  are  given,  the  cosine  of 
the  angle  is  the  number  by  which  to  divide  the  abscissa  to 
obtain  the  modulus. 

192 


Fig.  94. 


§  104]  THE   COSINE   FAMILY.  193 

LABORATORY  EXERCISES. 

Construct  with  the  protractor  an  angle  of  26°.  Lay  out  on  it  five 
different  moduli  and  corresponding  abscissas,  including  the  modulus  of 
unity  (one  inch  or  one  foot).     Measure  moduli  and  abscissas. 

Divide  each  abscissa  by  its  modulus  to  one  or  two  figures.  Compare 
results  with  one  another  and  with  the  table. 

Do  the  same  for  five  different  angles,  each  with  a  single  modulus  and 
its  abscissa.     Compare  results  with  the  tables. 

Lay  out  an  angle  with  the  protractor.  Measure  the  modulus.  Mul- 
tiply it  by  the  table-cosine  (to  one  or  two  figures).  See  if  the  result 
agrees  with  the  measured  abscissa. 

Test  for  another  angle  the  measured  modulus  with  the  value  obtained 
by  dividing  the  measured  abscissa  by  the  table-cosine. 

EXERCISES. 

1.  Take  the  exercises  under  §  76,  changing  the  word  a  sine  n  to  "  co- 
sine," and  "  ordinate  "  to  "  abscissa." 

2.  The  student  may  make  and  solve  five  simple  practical  examples 
which  can  be  solved  by  the  cosine ;  by  the  sine. 

3.  If  the  initial  velocity  of  a  projectile  at  an  angle  A°  to  the  hori- 
zontal is  v,  what  are  the  vertical  and  horizontal  components  ?  Solve  a 
numerical  example. 

4.  Assuming  that  the  velocity  due  to  gravity  is  32 1  (t  in  seconds), 
how  long  will  the  projectile  in  Ex.  3  rise  before  gravity  destroys  the 
initial  vertical  velocity?  How  far  will  the  projectile  drift  horizontally 
in  that  time  (range)  ? 

5.  A  smooth  plane,  inclined  A  °  to  the  vertical,  has  a  weight  of  \  pound 
tied  on  by  a  string  running  from  the  upper  end.  What  is  the  pressure 
normal  to  the  plane  ?  What  is  the  pull  on  the  string  ?  Solve  a  numerical 
example. 

6.  How  far  from  the  centre  of  the  earth  is  the  centre  of  the  40°  par- 
allel of  latitude?  What  is  the  radius  of  this  parallel?  Give  general 
formulas  for  sphere  of  radius  r. 

7.  A  particle  is  moving  with  uniform  angular  velocity  a  in  a  circle. 
What  are  the  component  velocities  along  horizontal  and  vertical  diameters 
at  the  time  t  (seconds),  counting  time  from  the  moment  when  the  particle 
is  at  the  right-hand  end  of  the  horizontal  diameter  ?  Solve  a  numerical 
example. 

8.  A  particle  is  acted  on  by  a  force,  F,  in  the  direction  A°  to  the 
horizontal  line.  What  are  the  horizontal  and  vertical  components  of  the 
force?     Solve  a  numerical  example. 


194 


PLANE   TRIGONOMETRY. 


[§104 


9.  A  particle  has  a  velocity  v  in  the  direction  A0  to  the  horizontal 
line.  What  are  the  horizontal  and  vertical  components?  Solve  a  numeri- 
cal example. 

10.   Compare  numerical  solutions  of  the  foregoing  problems  with  their 
graphical  solutions. 

Hassler's  definition  holds  for  all  positions 'of  the  terminal, 
no  matter  in  which  quadrant  it  may  fall. 


p 

I                                 5s 

X-                     -**"    ■ 

S 

S 

i  s 

^L* 

M 

_c 

Fig.  95. 

-M              O    - 

2 

JL 

_/ 

7 

7 

z 

/ 

7* 

\        n 

"S 

\^ 

^ 

S^ 

s^ 

^ 

]\ 

V. 

^ 

M                     0 

Fig.  96. 

^                     -.*- 

0                      M 

^^ 

v. 

nn. 

^v 

i                                                 ^£ 

£     ^ 

Fig.  97. 


Fig.  98. 


Representing  by  0  any  one  of  the  angles  corresponding  to 
the  terminal  in  any  one  of  the  above  diagrams, 

£  n      abscissa  OM     x  n 

cosine  of  o  == — —  =  -  =  cos  v. 

modulus  OP     r 


106] 


THE   COSINE  FAMILY. 


195 


§  105.    The  Sign  of  the  Cosine. 


The  sign  of  the  cosine  is  deter- 
mined by  that  of  the  abscissa,  as 
that  of  the  sine  by  the  ordinate. 


Quadrant 

I 

E 

K 

K 

Cosine 

-h 

— 

— 

+■ 

Sine 

+ 

+ 

— 

— 

Fig.  99. 


EXERCISES. 

1.  By  means  of  the  protractor,  locate  the  terminals  of  the  following 
angles  (in  degree  measure).  Determine  the  signs  of  their  sines  and 
cosines:  30;  -30;  60;  -60;  110;  -110;  130;  -130;  271;  -271; 
375;  -375;  456;   -456;   548;   -548;   638;   -638. 

Do  you  notice  any  connection  in  size  and  sign  between  the  cosine  of 
positive  and  negative  angles  numerically  equal?  Can  the  sign  of  the 
sine  and  cosine  be  determined  without  knowing  accurately  the  position 
of  the  terminal  ? 

2.  Give  the  signs  of  the  sines  and  cosines  of  the  following  angles 
(radian  measure),  taking  each  angle  both  positively  and  negatively: 


1;    2;    3;    4;    5;   7;     9;    10;    12;    ?;    ?;    £;    £2 

6      o      4       o 


T'  T'  T' 


7tT         9tT.      13  7T 


3tt 

2  ' 


§  106.    Angles  with  the  Same  Cosine. 

The  cosine,  like  the  sine,  depends  only  on  the  position  of 
the  terminal. 

(i)  All  angles  with  the  same  terminal 
have  the  same  cosine. 
.\  cos  (2  mr  -f-  0)  =  cos  0, 

cos  (2  n  •  180°  +  A°)  =  cos  A°, 
where   n   is   any   positive   or   negative 
integer. 


Fig.  100. 


(ii)  All  angles  with  their  terminals  symmetric  to  the 
horizontal  have  the  same  cosine,  since  points  with  equal 
moduli  on  such  terminals  have  the  same  abscissa. 


196 


PLANE   TRIGONOMETRY. 


[§106 


Fig.  101. 


.*.  as  a  special  case, 

cos  6  =  cos  (—  #),  in  radian  measure, 
cos  A°  =  cos  (—J.0),  in  degree  measure, 

or,  positive  and  negative  angles,  numeri- 
cally equal  have  the  same  cosine. 

And  generally, 
cos  (2  rnr  +  0)  =  cos  (2  m  •  ir  —  0), 
cos(2n-180o  +  A°)  =  cos(2m.l80o-A°), 

where  m,  n  are  any  positive  or  negative 
integers. 

(iii)  Thus,  in  general,  any  angle 
having  the  same  cosine  as  0,  A°  are  of 
the  forms 

2  nir  ±  0,  in  radian  measure, 

2  n  •  180°  ±  A°,  in  degree  measure, 

where  n  is  any  positive  or  negative  integer. 

The  student  may  give  a  verbal  statement  for  each  of  these 
formulas. 

(iv)  In  the  preceding  formulas  any  angle  of  the  given 
terminal  may  be  taken  as  0,  A0.  It  is  customary,  however, 
to  take  the  principal  angle.  For  example,  185°  and  — 175° 
locate  the  same  terminal.     One  would  use 

2rc- 180°  ±175°, 

rather  than  2  n  •  180°  ±  185°, 

for  the  set  of  angles  having  the  cosine  of  this  terminal. 


Fig.  102. 


EXERCISES. 

1.  Find  five  positive  angles  and  five  negative  angles  having  the  same 
cosine  as  30°,  and  also  give  in  degree  measure  and  in  radian  measure  the 
general  formulas  for  all  such  angles. 

2 .  Do  the  same,  replacing  the  word  "  cosine  "  by  the  word  "  sine."  Give 
a  diagram  for  Exs.  1  and  2. 


§  106]  THE   COSINE   FAMILY.  197 

3.  Consider  in  the  same  way  as  in  Exs.  1  and  2,  the  angles  100°,  193°, 
275°,  using  the  principal  angle  of  the  terminals. 

4.  For  terminals  in  the  first  quadrant,  where  are  the  terminals  for  all 
angles  having  the  same  cosine  ?  For  terminals  in  the  second  quadrant  ? 
In  the  third  ?    In  the  fourth  ? 

5.  Give  the  general  formulas,  both  in  radian  measure  and  in4§Are0^ 

measure,  for  all  angles  having  the  same  cosine  as  -;  — •    — .   2JF.  UL. 
r  8  8  4'   3'     3  '     4  '     4  ' 

-;  —  •     For  the  same  angles  give  the  general  formulas,  in  both  meas- 
6       6 
ures,  for  all  angles  having  the  same  sine.     Give  illustrating  diagrams. 

6.  Find  the  two  angles  numerically  less  than  180°  which  have  the  same 
cosine  as  ±  1085°.  Do  the  same  for  ±  365°;  ±  800°  ;  ±  1100°;  ±  3000°; 
±  185° ;  ±  275°.  Give  in  radian  measure  and  in  degree  measure,  for  each 
of  these  angles,  the  general  formulas  for  all  angles  having  the  same  cosines. 

7.  Consider  Ex.  6,  replacing  the  word  " cosine "  by  the  word  "sine." 

8.  Using  tt  radians  as  the  unit  angle,  give  the  general  formulas  for 
Exs.  5  and  6. 

9.  Solve  the  equations  : 

cos  9  6  =  cos  8  0 ;        cos  7  0  =  cos  6 ;      —  cos  3  0  =  cos  5  $ ; 
cos  2  0  =  cos  0 ;       —  cos  3  0  =  cos  0 ;  cos  n$  =  cos  0 ; 

cos  md  =  cos  nO ;         sin  2  0  =  sin  6  0 ;  —  sin  8  $  =  sin  7  6 ; 
sin  md  =  sin  n9. 
Illustrate  each  solution  by  a  diagram. 

Solution  of  the  first  equation  in  Ex.  9. 

cos  9  6  =  cos  8  6. 

Here  9  6  is  not  necessarily  8  6,  but  9  6  may  be  any  one  of 
the  angles  whose  cosine  is  that  of  8  0. 

But                     cos  8  6  =  cos  (2  mr  ±  8  0). 
.-.  cos  9  6  =  cos  (2  nir  ±  8  0). 
.-.  9  6  =  2  tot  +  8  0,  or 
9  0  m  2  mr  -  8  0. 
.-.  0  =  2  nir  ;              (1) 
or,  17  0  =  2  nir  ;  (2) Termtnal 

or,  0  =  n-— .  (3)  FlG-103'   (•«*«*•> 

17 


198 


PLANE   TRIGONOMETRY. 


[§106 


80 


Jfermfi 


jertft 


Fig.  104. 


('-*■) 


For  (1)  the  terminal  line  is  the  initial  line.  Turning  that 
line  nine  times  or  eight  times  around  leaves  it  unchanged. 

The  diagram  is  Fig.  103. 
J9_  For  (3), 

9  (9 is  9  n .  t2ytt,  or  n(l-T8y)2  tt; 
8 0  is  8  n  •  &  w,  or  n(l  -  T9y)2  tt. 

Thus  the  terminal  for  9  6  is 
symmetric  to  the  horizontal 
with  that  of  8(9.  The  dia- 
gram for  n  =  1  is  Fig.  104. 

The  student  may  give  similar  solutions  for  the  remaining  examples, 
and  state  what  the  general  process  is. 

§  107.     Angles  with  Opposite  Cosines. 

(a)  Any  pair  of  terminals  symmetric  to  the  vertical  give 
angles  with  opposite  cosines,  since  for  equal  moduli  the  ab- 
scissas are  oppositely  equal. 

.\  (i)  Supplementary  angles  have 
opposite  cosines, 

or,        cos  (180°  -  A°)  =  -  cos  A0, 
cos  (tt  —  0)  =  —  cos  6. 

(ii)  Any  angle  of  a  terminal  has  the 
opposite  cosine  of  any  angle  of  the  ter- 
minal symmetric  to  the  vertical.  Fig.  105. 

cos  {  (2  n  +  1)  tt  -  6 }  =  -  cos  (2  m  •  tt  +  0), 
cos  S  (2  n  +  1)  180°  -  A0 \  =  -  cos  (2  m  .  180°  +  A°). 
(iii)  Thus  all  angles  of  the  form 
(2  n  +  1)  tt  -  (9, 
(2rc  +  l)180°-^°, 

have  the  opposite  cosine  of  0,  A°. 

In  words,  subtracting  an  angle  from 
an  odd  multiple  of  180°  (or  tt  radians} 
Fig.  106.  gives  an  angle  of  opposite  cosine. 


M. 


V 


Ml 


107] 


THE   COSINE  FAMILY. 


199 


jr. 


(#)  Angles  of  opposite  terminals 
have  opposite  cosines,  since  for  equal 
moduli  the  abscissas  are  oppositely- 
equal. 

.\  cos(180°  +  ^Lo)  =  -cosJ.°, 

cos  (tt  -f-  0)  =  —  COS  #, 

or,  in  general,  Fia.  107. 

cos  (  (2  n  +  1)  180°  +  A0)  =  -  cos  (2  m  •  180°  +  A°), 
cos((2tt+l)?r  +  0)=  -cos  (2  wtt  +  0). 

In  general,  angles  having  the  oppo- 
site cosine  of  0,  A°  are  of  the  form 
(2n  +  l)7r±!9, 

(2  7i  +  1)180°  ±A°. 

Thus,  while  adding  an  angle  to  an 
even  multiple  of  180°  or  subtracting  it 
from   such   an  angle  leaves  the  cosine 

Fio  108 

unchanged    (§  106),    an  odd    multiple 

of  180°,  similarly  treated,  gives  the  opposite  cosine. 


EXERCISES. 

In  solving  examples,  it  is  advised  that  the  formulas  be  not  used,  but 
rather  the  location  of  the  terminal,  as  with  the  sine. 

1.  Write  five  positive  angles  and  five  negative  angles  having  the  oppo- 
site cosine  of  30°;  of  ±  60°;  of  ±  45°.  Let  one-half  the  angles  found  in 
each  case  have  the  opposite  terminal,  the  remaining  half  the  terminal 
symmetric  to  the  vertical.     Give  the  general  formulas  in  each  case. 


2.   Do  the  same  for 


-2tt.    3tt 


3'    6'     3 


;   2";   3";  4-. 


3.  If  an  angle  has  its  terminal  in  the  first  quadrant,  in  which  quadrants 
will  the  angles  of  opposite  cosines  have  their  terminals  ?  Consider  each 
quadrant  similarly. 


4.   Change  the  word  "  cosines  "  in  Ex.  3  to  "  sines,"  and  solve. 


200  PLANE   TRIGONOMETRY.  [§  108 


§  108.    On  the  Effect  on  a  Diagram  of  a  Quarter-turn,  Half- 
turn,  Three-quarter-turn,  Counter-clockwise. 

By  a  "quarter-turn  counter-clockwise"  is  meant  a  tilt 
through  90°  counter-clockwise.  Evidently  after  such  a 
tilt: 

A  line  now  vertically  up  was  originally  right-flat. 

A  line  now  right-flat  was  originally  vertical-down. 

Or,  a  line  now  a  positive  ordinate  was  a  positive  abscissa. 

A  line  now  a  positive  abscissa  was  a  negative  ordinate. 

To  the  word  "  ordinate  "  we  may  consider  the  word  "  sine  " 
attached ;  to  positive  ordinate,  positive  sine  ;  to  negative 
ordinate,  negative  sine.  The  word  "  cosine  "  we  attach  simi- 
larly to  the  word  "abscissa." 

Thus,  "  a  positive  ordinate  was  a  positive  abscissa,"  is  the 
same  as  "  a  positive  sine  was  a  positive  cosine,"  or,  "  a  sine 
was  a  cosine";  while  "a  positive  abscissa  was  a  negative 
ordinate"  means  "a  cosine  was  a  negative  sine." 

.-.  sin  (90°  +  ^4°)  =  cos  A° ;  cos  (90°  +  AT)  »—  sin  A°. 

Similarly,  in  considering  any  question  of  this  kind,  it  is 
necessary  only  to  determine  where  the  ordinate,  upright  after 
the  tilt,  was  before  the  tilt,  and  where  the  abscissa,  flat  to  the 
right  after  the  tilt,  was  before  the  tilt. 

What  effect  has  a  half-turn  counter-clockwise  ? 

A  line  up,  was  down;  right-flat  was  left-flat. 

Thus,  a  positive  ordinate  was  a  negative  ordinate,  and  a 
positive  abscissa  was  a  negative  abscissa. 

.-.  sin  (180°  +  A°)  m  -  sin  A° ;  cos  (180°  +  A0)  =  -  cos  A°. 

What  effect  has  a  three-quarter  counter-clockwise  turn? 
A  positive  ordinate  was  a  negative  abscissa,  and  a  positive 
abscissa  was  a  positive  ordinate. 

.-.  sin(270°-h^o)  =  -cos^L°';  cos  (270°  +  A0)  =  sin  A0. 


§  108]  THE   COSINE   FAMILY.  201 

What  effect  has  a  complete  turn  ? 
The  diagram  is  unaffected. 

.\  sin  (360° +  .4°)  =  sin  ^°;  cos  (360°  +  4°)  =  cosui°. 

What  effect  on  a  diagram  has  any  number  of  quarter-turns, 
counter-clockwise  t 

From  all  such  tilts  all  complete  turns  may  be  dropped,  as  be- 
ing without  effect.  There  will  remain  a  quarter-turn,  a  half- 
turn,  or  a  three-quarter-turn.    The  corresponding  angles  are  : 

4 rc.90°,  (4ti  +  1)90°,  (4rc  +  2)90°,  (4ti  +  3)90°. 

f  sin  (4  n.90° +  ^°)  =  sin  ^1°.     | 

{  cos(47i.90°  +  7l°)  =  cosJ.°.     J 

|  sin  (  (4  7i +  1)90°  + ^4°)  =  cos  ^1°.     \ 
(cos((47i  +  l)90°  +  ^°)  =  -sinA°.J 

f  sin ( (4  7i  +  2) 90°  +  A°)  =  -  sin^4°.  | 
{  cos( (4 n  +  2)90°  +  4°)  =  -  cos^4°.  J 

f  sin((47i  +  3)90°  +  A°)  =  -cos^4°.| 
{cos  ((4  7i  +  3)90°  +  ^l°)  =  sin  ^4°.     J 

If  B,  <7,  D,  are  three  angles  in  degree  measure  free  from 
multiples  of  360°,  and  of  the  second,  third,  fourth  quadrant, 
respectively,  then  £-90°,  (7-180°,  B  -  270°,  are  three 
angles,  each  less  than  90°,  whose  terminals,  by  a  quarter- 
turn,  half-turn,  three-quarter-turn,  respectively,  come  into 
coincidence  with  those  of  B,  C,  D,  respectively.     Thus, 

«) 


JsinB=     cos(J5-90°).   } 
|cosB  =  -sin(B-90°).   J 

(sinC  =  -sin(C-180°).K... 
{ cos  C  =  -  cos  (C  -  180°). ) w 

f  sin  D  =  -  cos  (Z>  -  270°). )  ( . . .. 
[cosI>  =     sin(i>  -  270°). )  K     J 


Formulas  (i),  (ii),  (Hi),  are  of  great  practical  importance  in  using  the 
tables,  obviating  subtractions  on  minutes  and  seconds.  The  teacher  may 
assign  practice  exercises. 


202  PLANE  TRIGONOMETRY.  [§  109 

EXERCISES. 

1.  Cut  out  a  right-triangle  from  paper,  carry  out  the  tilts  indicated 
in  §  108,  and  thus  prove  the  formulas. 

2.  Determine  what  the  formulas  above  become  when  n  is  a  negative 
integer. 

3.  Change  A  to  —  A  in  the  formulas  of  this  section,  and  from  the 
facts  sin  A  =  —  sin  (—  A),  cos  A  =  cos  (—  A),  deduce  the  values  of  the 
following  expressions  in  terms  of  A:  sin  (90°  —  A);  cos  (90°  —  A) 
sin  (180°  -  A)  ;  cos  (180°  -  A)  ;  sin  (270°  -  A )  ;  cos  (270°  -  A) 
sin  (360°  -  A)  ;  cos  (360°-  A)  ;  sin  (4n-  90°  -  A)  ;  cos  (4n  •  90°-  A) 
sin  ((4  n  +  1)90°  -  A)  ;  cos  ((4  n  +  1)90°  -  A)  ;  sin  ((4 n  +  2)90°  -  A) 
cos  ((4  n  +  2)90°  -  A)  ;  sin  ( (4  n  +  3)90°  -  A)  ;  cos  ((4  n  +  3)90°  -4). 

4.  Solve  Ex.  3  by  diagrams. 

5.  Determine  the  values  of  the  expressions  in  Ex.  3  when  —  90°  is 
written  for  90°. 

6.  Construct  diagrams  for  the  expressions  90°  ±  A,  etc.,  of  this  section 
and  the  preceding  examples,  and  show  the  results  directly  from  the 
diagrams. 

§  109.  The  following  generalization  may  be  made  from  the 
preceding  results  : 

If  the  subtracted  (added)  90°'s  are  even,  the  word  "  sine  " 
remains  the  word  "  sme,"  and  the  word  "  cosine  "  remains  the 
word  "cosine."  When,  however,  their  number  is  odd,  the  word 
"sine"  changes  to  "cosine"  and  the  word  "cosine"  to  "sine." 
The  proper  sign  is  then  to  be  attached,  according  to  the  relative 
position  of  the  original  terminal  and  the  new  terminal. 

EXERCISE. 

Determine  by  the  method  of  tilts  and  by  the  preceding  suggestion,  the 
effect  on  the  sine  and  cosine  of  adding  (subtracting)  the  first  fifty  multi- 
ples of  90°  to  an  angle  A,  taking  A  in  each  quadrant. 

For  example,  what  is  the  effect  of  adding  twenty-seven 
90°'s  to  the  angle  123°  ? 

Dividing  27  by  4,  the  remainder  is  3.  The  effect,  then, 
is  the  same  as  that  of  a  three-quarter  tilt,  counter-clockwise. 


§  110]  THE   COSINE   FAMILY.  203 

The  positive  ordinate  was  the  negative  abscissa. 

The  positive  abscissa  was  the  positive  ordinate. 

.-.  sine  was  a  negative  cosine,  and  cosine  was  a  sine. 

.-.  sin  (27  x  90°  +  123°)  =  -  cos  123°, 
and  cos  (27  x  90°  +  123°)  =  sin  123°. 

Using  the  second  method  : 

Since  27  is  an  odd  number,  the  words  interchange,  "  sine  " 
becoming  "cosine"  and  "cosine,"  "sine."  Omitting  from 
27  all  multiples  of  4,  the  remainder  is  3.  The  angle  123°  is 
of  the  second  quadrant.  Adding  three  90°'s  will  bring 
the  new  terminal  to  the  first  quadrant.  A  sine  in  the  first 
quadrant  is  opposite  in  sign  to  a  cosine  of  the  second. 

.\  sin  (27  x  90°  +  123°)  =  -  cos  123°. 

But  a  cosine  of  the  first  quadrant  is  of  the  same  sign  as  a 
sine  of  the  second. 

cos  (27  x  90°  +  123°)  =  sin  123°. 

I  prefer  the  method  of  tilts.  The  student  may  take  his 
choice,  or  devise  a  better  method. 


§  110.    Cosines  of  all  Angles  are  First  Quadrant  Cosines. 

It  is  evident  that  if  the  cosines  of  all  angles  of  the  first 
quadrant,  between  0°  and  90°,  are  known,  the  cosines  of  all 
other  angles  are  known  in  magnitude. 

For  a  terminal  of  quadrant  II,  use  the  symmetric  terminal 
of  quadrant  I,  with  change  of  sign  of  the  cosine. 

For  a  terminal  of  quadrant  III,  use  the  opposite  terminal 
of  quadrant  I,  with  change  of  sign  of  the  cosine. 

For  terminal  of  quadrant  IV,  use  the  symmetric  terminal 
of  quadrant  I,  without  change  of  sign  of  the  cosine. 


204  PLANE   TRIGONOMETRY.  [§110 

Take  in  each  case  the  principal  angle  of  the  corresponding 
terminal  in  quadrant  I. 

Without  constructing  the  terminal  one  may  proceed 
thus  : 

(a)  Disregard  the  sign  of  the  given  angle,  since  cos  (  —  A) 
=  cos  A. 
(5)  Drop  all  multiples  of  360°,  as  being  without  effect. 

0)  For    remainder,    R,    if    R  >  90°    and    <  180°,   take 

-  cos  (180°  -  R),  or  -  sin  (R  -  90°)  (§  108). 

(d)  For  remainder,  R,  if  R  >  180°  and  <  270°,  take 
-cos  (5-  180°), 

O)  For  remainder,  R,  if  R  >  270°  and  <  360°,  take 
cos  (360°  -  E),  or  cos(i2  -  270°)  (§  108). 

The  student  will  observe,  on  comparing  these  statements 
with  terminal  diagrams,  that  they  are  nothing  but : 

Terminals  symmetric  to  the  vertical,  give  opposite  cosines. 
Terminals  opposite,  give  opposite  cosines. 
Terminals   symmetric   to   the   horizontal,  give  the  same 
cosines. 

EXERCISES. 

1.  Give  the  angles  between  0°  and  90°  which  have  the  same  cosine  in 
numerical  magnitude  as  the  following  degree-measure  angles,  and  give 
the  sign  in  each  case,  this  being  obtained  from  the  quadrant  of  the  given 
angle : 

140°;     175°;     187°;     200°;    280°;    305°  23';    -30°  23';     -100°  32'; 

-  185°  43' ;    -  5°  43' ;    -  150°  25' ;    ±  432° ;    ±  403°  ;    ±  506° ;    ±  603° ; 
±1850°;    ±2317°. 

2.  Solve  the  preceding  when  for  the  word  "  cosine  "  there  is  read  the 
word  "  sine." 

3.  Express  two  of  the  preceding  angles  in  radian  measure,  (i)  in 
terms  of  a  radian,  (ii)  in  terms  of  it  radians. 

4.  Solve  Ex.  1  by  constructing  terminals. 


§111] 


THE   COSINE  FAMILY. 


205 


§  111.   Construction  of  the  Terminals  of  Angles  having  a  Given 

Cosine. 

(i)  Let  cos  A  =  § . 

Draw  a  circle  with  the  radius  3.  Lay  out  on  the  abscissa 
axis  the  distance  OM=  2 ;  then  draw  through  M  the  verti- 
cal line,  cutting  the  circle  in  Px 
P4.  OPv  OP±  are  the  required 
terminals. 

For  every  given  value  of  the 
cosine,  other  than  ±  1,  there  are 
two  and  only  two  positions  of 
the  terminal.  These  two  posi- 
tions are  always  symmetric  to 
the  horizontal,  as  were  terminals 
for  a  given  sine  symmetric  to 
the  vertical  (§  66).  fig.  109. 

The  angles  of  the  two  preceding  terminals  are 

2w.l80°±JL°(§106,  iii). 

(ii)  Let  cos  A  =  —  J. 

The  construction  is  the  same  as  that  above,  except  that 
OM  is  laid  out  to  the  left,  or  OM  =  -  2. 

EXERCISES. 


1.  Construct  terminals  for  the  cosines  £,  |,  |,  I,  and  give  the  general 
value  of  the  angles,  using  tables.     Test  with  protractor. 

2.  Show  that  taking  some  line  for  unity,  one  can  construct  lengths 

representing  the  square  roots  of  2,  3,  5,  6,  7,  8,  or  any  other  integer. 

Use  the  process  to  construct  the  terminals  for  the  angles  whose  cosines 

are+V2     ,V3       V5 
are  ±_ ,  ±— ,  ±— 

3.  Can  you  determine  from  the  constructions,  without  measurement 
or  tables,  what  angles  have  the  cosines  ± — ,  ± — ,  ±1,  ±1,  0? 

4.  Change  the  word  "cosines"  to  "sines"  in  Exs.  1,  2,  and  3,  and  solve. 

5.  Construct  terminals  for  some  cosines  in  the  tables.    Measure  the 
angle  with  the  protractor  and  compare  with  the  table. 


206 


PLANE   TRIGONOMETRY. 


[§112 


§  112.    Line  Picture  of  the  Cosine. 


::zj::-:-:-::::_::: 

~k^f                                           ^        D 

tit            ......          ^ 

~$             ~        _^z    s 

*T-                            7          v 

5                -    y         X 

h  •-■■■■                    Z                           Ij! 

2 

O    -             -ifef 

t                                         H    4 

A                                              A 

\         -*i                                           V 

5                                      ^Z 

^                                   Z 

\      »_                       ^ 

--J — - 

Fig.  110. 


As  on  the  unit-circle  the  ordi- 
nates  represent  the  sines  for  the 
corresponding  terminals,  so  do 
the  abscissas  represent  cosines. 

For  the  terminal  OP,  the 
cosine  is 

This  is  the  line  definition  of 
the  cosine.     Compare  §  67. 


LABORATORY  EXERCISE. 

Divide  a  circumference  whose  radius  is  one  foot  into  5°  spaces ;  measure 
the  abscissas ;  compare  with  the  table-cosines. 


§  113.    Line  Pictures  of  the  Cosines  of  All  Angles. 

As  all  the  ordinates  of  the  unit-circle  represent,  as  a 
continuum,  the  sines  of  all  angles  (see  §  69),  so  do  the 
abscissas  give,  pictorially,  a  cor- 
rect impression  of  the  relative 
magnitude  of  the  cosines  of  all 
angles,  with  their  manner  and  ex- 
tent of  variation  as  the  terminal 
passes  from  the  initial  position 
OA,  counter-clockwise,  through 
a  perigon,  and  continues  its  mo- 
tion, with  a  repetition  of  the 
cosine  values. 

From  the  diagram  (Fig.  Ill) 
one  can  readily  draw  the  follow- 
ing conclusions : 

(a)  When  the  terminal  is  horizontal  to  the  right,  the  cosine 
is  +1. 


Fig.  ill. 


§113d] 


THE   COSINE   FAMILY. 


207 


.-.  cos0°=l  =  cos(±360o)=cos(±720°),  etc., 
or,  in  general, 

cos  (2  n  •  180°)  =  1  =  cos  (2  n  •  tt), 
where  n  is  any  ±  integer. 

That  is,  the  cosine  of  all  even  multiples  of  180°,  or  of  ir 
radians,  is  +1. 

(5)  When  the  terminal  is  horizontal  to  the  left,  the  cosine 
is  —  1. 

.\  cos  (  ±  180°)  =  -  1  =  cos  (  ±  540°),  etc., 

or,  in  general, 

cos(2  n  +  1)180°  =  -  1  =  cos(2  n  +  1)tt. 

What  is  the  cos  (  ±  ir)  ? 

The  cosine  of  all  odd  multiples  of  180°,  or  of  ir  radians,  is 
-1. 

(c)  When  the  terminal  is  vertical,  up  or  down,  the  cosine  is 
zero. 

.-.  cos  (  ±  90°)  =  0  =  cos  (  ±  270°)  =  cos  (  ±  540°),  etc. 

cos(  ±  ^j=  0=  cos(  ±  -£-)=  cos(  ±-^j,  etc. 

In  general, 

cos(2w  +  l)90°  =  0  =  cos(2w  +  l)^- 

The  cosine  of  all  odd  multiples  of  90°,  or  of  —  radians,  is 
zero. 

((i)   Thus,  for  terminals  bordering  the  quadrants : 


Terminal 

Right 

Up 

Left 

Down 

Right 

Cosine 

+  1 

0 

-1 

0 

+  1 

Sine 

0 

+1 

0 

-1 

0 

Fia.  112. 


EXERCISE. 

Name  five  positive  angles  and  five  negative  angles  in  degree  measure 
and  in  radian  measure  for  each  terminal  corresponding  to  Fig.  112. 


208 


PLANE  TRIGONOMETRY. 


[§  113  e 


(e)  As  the  terminal  passes  from  0°  to  360°,  the  cosine, 
starting  with  the  value  +  1,  diminishes  continuously  from 
1  to  0,  which  is  reached  at  90°;  then  diminishes  continu- 
ously to  —  1,  which  is  reached  at  180° ;  then  increases  con- 
tinuously to  0,  which  is  reached  at  270°;  then  increases 
continuously  to  -f-  1,  which  is  reached  at  360°.  As  the  ter- 
minal passes  again  around  the  circuit,  the  same  set  of  values 
is  repeated,  and  so  on.  Thus,  as  the  angles  form  a  continuum 
from  0°  to  360°,  the  cosines  form  a  double  continuum  from 
+  1  to  -  1,  and  -  1  to  + 1. 

0°  90°  180°         270°        360 


+  1 


The  Cosine. 


The  Sine. 


-l 


+  i 


-l 


^ ^ 

ST 

^V^^ 


Fig.  113. 

90°         180° 


270° 


360' 


^ar 

"^ 

Sk 

Repeating,  forward 
and  backward, 
without  end. 

Repeating,  forward 
and  backward, 
without  end. 


Fig.  114. 


EXERCISE. 

What  effect  has  a  reversal  of  the  order  of  angle  description  on  the  order 
of  description  of  the  double  cosine-continuum  ?    On  the  sine-continuum  ? 

(/)  For  any  position  on  the  terminal,  a  slight  change  in  the  ter- 
minal position  makes  a  slight  change  in  the  corresponding  cosine. 
In  Fig.  115,  OMv  OM2  are  the  cosines  of  two  angles  nearly 

equal.  The  cosine-difference  is 
M1M2.  As  P2  is  brought  nearer 
and  nearer  to  Pv  the  smaller 
becomes  M1Mi.  When  the  angle- 
difference  is  less  than  any  as- 
signable quantity,  so  is  the 
cosine-difference,  for  M^M2  is  less 


Fig.  115. 


than  PXP2,  this  arc-difference  on 


the  unit-circle  being  the  same,  numerically,  as  the  angle- 
difference  in  radian  measure.     See  (4),  page  67. 

Thus  the  cosine  is  a  continuous  function  of  the  angle. 


§  113  i]  THE   COSINE  FAMILY.  209 

(#)  For  all  angles  whose  terminals  are  in  the  first  or 
second  quadrant,  increasing  the  angle  diminishes  the  cosine. 

Therefore  all  angles  with  positive  sines  have  for  an  in- 
crease in  the  angle  a  decrease  in  the  cosine. 

Similarly,  considering  the  third  and  fourth  quadrants,  all 
angles  with  negative  sines  have,  for  an  increase  in  the  angle, 
an  increase  in  the  cosine. 

(Ji)  The  points  at  which  a  function  changes  from  an  in- 
creasing function  to  a  decreasing  function,  or  vice  versa,  are 
called  the  turning-points  of  the  function. 

EXERCISES. 

1.  In  what  position  is  the  terminal  when  the  cosine  is  at  a  turning- 
point  ? 

2.  When  the  sine  is  at  a  turning-point? 

3.  Give  in  circular  measure  and  in  degree  measure  general  expres- 
sions for  all  angles  whose  cosines  are  at  a  turning-point.  Do  the  same 
for  the  sine. 

(i)  When  the  terminal  is  nearly  flat,  a  slight  change  hi  its 
position  does  not  affect  the  cosine  very  materially.  When,  how- 
ever, the  terminal  is  very  nearly  vertical,  slight  changes  in  its 
position  are  accompanied  by  marked  changes  in  the  cosine. 

From  this  it  follows  that  small  angles  cannot  be  deter- 
mined with  great  accuracy  from  the  cosine,  just  as  angles 
near  90°  cannot  be  determined  accurately  from  the  sine.  In 
practical  work,  it  is  well  to  avoid,  when  possible,  observations 
which  lead  to  the  determination  of  an  angle  from  its  sine, 
when  the  angle  is  within  5  or  6  degrees  of  90°,  or  from  its 
cosine,  when  the  angle  is  within  5  or  6  degrees  of  zero. 
Information  in  detail  will  be  given  on  this  point  later.  See 
§§  185,  186. 

EXERCISE. 

Examine  four-place  tables  and  five-place  tables  for  the  changes  made  in 
the  sine  and  cosine  for  a  change  of  1'  in  the  angle  at  different  parts  of 
the  tables.    Examine  also  the  corresponding  changes  in  log  sin,  log  cos. 


210  PLANE   TRIGONOMETRY.  [§  114 

§  114.     The  Cosine  as  a  Function  of  the  Angle. 

The  relation  between  the  cosine  and  the  corresponding 
angle  in  radian  measure  is,  as  will  be  shown  later, 

cos0  =  l--  +  ---+..,etc. 

Compare  this  with  the  sine  series,  or  sine  function,  as  given 
in  §  70. 

The  cosine,  like  the  sine,  is  thus  a  transcendental  function 
of  the  angle. 

While  the  sine  involves  only  odd  powers  of  the  angle, 
the  cosine  is  expressed  wholly  in  terms  of  the  even  powers. 
This  is  essential,  since 

sin(<9)=-sin(-0), 
and  cos(0)=  cos(—  0). 

The  cosine  is  thus  an  even  function  of  the  angle,  while  the 
sine  it  an  odd  function  of  the  angle. 

EXERCISES. 

1.  Show  from  the  function  expressions  for  sine  and  cosine  that 
sin  (0)  =  —  sin  (  —  0)  and  cos  ($)  =  cos  (  —  $). 

2.  Show  in  general  that  if  any  function  is  expressed  in  powers  of  x 
and  if  F(x)  =  F  (—  x),  there  can  be  only  even  powers  of  x,  while  if 
F  (x)  =  —  F  (—  x),  there  can  be  only  odd  powers  of  x. 

3.  Use  the  sine  and  cosine  function  expressions  to  calculate  the  sines 
and  cosines,  to  two  significant  figures,  of  the  following  angles : 

0,  1",  1',  1°,  ±  2°,  ±  3°,  ±  5°,  1',  2Tt  Ti-. 


§  115.   The  Cosine  as  a  Periodic  Function. 

Since  cos  6  =  cos  (6  -f-  n  •  2  7r), 

the  cosine  is,  like  the  sine,  periodic,  with  the  period,  2  7r. 
Reread  §  72. 


116]  THE  COSINE  FAMILY.  211 


116.     The  Anti-cosine,  or  Inverse-cosine. 


When  y  =  cos  x,  then  x  =  cos-1  y . 
Reread  §  73. 


EXERCISES. 

1.  Read  in  all  possible  ways,  as  in  the  corresponding  exercise  on  the 
sine,  the  expressions : 

cos-1 1;  cos-1 1;  cos"1  — ;  cos-1-—;  sin-1^;  sin-1  ( );  sin  (cos-1  £); 

cos  (sin-1  0);       cos  (cos-1  x)  ;        sin  (cos-1  0)  ;        3  sin-1  \  +  2  cos-1 — ; 
3  sin  (cos-1  \)  -  2  cos  (sin-1  f  ).  2 

2.  Construct  the  terminals  for  such  of  the  expressions  in  Ex.  1  as 
represent  angles. 

3.  How  many  values  have  the  following  expressions? 

sin  (sin-1  x)  ;  sin  (cos-1  a;)  ;  cos  (cos-1  x)  ;  cos  (sin-1  x). 
Read  these  expressions  in  all  possible  ways. 

4.  How  many  values  have  the  following  expressions  ? 

cos-1  (cos  x)  ;  sin-1  (sin  x)  ;  sin-1  (cos  x)  ;  cos-1  (sin  x). 
Read  these  expressions  in  all  possible  ways. 

5.  What  are  the  values  of  the  following  expressions  ? 

sin  (cos-1  0)  ;         sin  (cos-1 1)  ;        sin  (cos-1  (  —  1))  ;        sin  (  —  cos-1 1)  ; 
sin  (—  cos-1  (—  1))  ;     sin  (—  cos-1  0). 

6.  Give  the  values  of  the  expressions  in  Ex.  5  when  the  words  "  sine  " 
and  "cosine"  are  interchanged.  When  both  words  are  "sine."  When 
both  words  are  "cosine." 

7.  Give  two  special  values  and  the  general  values  of  each  of  the 
following  expressions: 

sin-1  (cos 0°);  sin"1  (cos  (±90°));  shr1  (cos  (±180°));  sin"1  (cos  (±270°)); 
sin"1  cos  (±360°)). 

8.  Interchange  the  words  "  sine  "  and  "  cosine  "  in  Ex.  7  and  solve. 

9.  Also  answer  Ex.  7  when  the  word  "  cosine  "  is  preceded  by  the  nega- 
tive sign.     Then  interchange  the  word  "  sine  "  and  "  cosine,"  and  answer. 


212 


PLANE  TRIGONOMETRY. 


[§117 


§  117.     Some  Angles  whose  Cosines  can  be  determined  readily 
from  a  Diagram. 

These  are  the  same  angles  for  which  the  sines  were  deter- 
mined in  §  75.  The  diagrams  used  for  sines  give  also  the 
cosines.     These  diagrams  are: 

For  45°  and  all  For  60°  and  all  For  30°  and  all 

angles  of  the  form     angles  of  the  form     angles  of  the  form 


2.    T 
nir  ±-7 

4 

2  n  •  180°  ±  45c 


2  nir  ± 


3 


2  n  ■  180°  ±  60c 


2nir±l 
6 

2  n  •  180°  ±  30c 


Fig.  116. 


Fig.  117. 


Fig.  118. 


The  cosines  of  all  angles  noted  in  §  75  are  readily  obtained 
from  these  diagrams  by  observing  whether,  as  related  to  the 
diagram  terminal,  the  new  terminal  is 

0)  opposite, 

(ii)  symmetric  to  the  horizontal, 

(iii)  symmetric  to  the  vertical,  remembering: 

at        -.lt_  -.  •     t    t         f  opposite  sines, 

Angles  with  opposite  terminals  have  i  ., 

&  ri  I  opposite  cosines. 

Angles  with  terminals  symmetric  to  the  [opposite  sines, 

horizontal  have  I  same  cosine. 

Angles  with  terminals  symmetric  to  the  rsame  sine, 

vertical  have   [opposite  cosines. 


§117] 


THE  COSINE   FAMILY. 


213 


EXERCISES. 

1.  Name  five  angles  having  their  terminals  opposite  to  that  of  45°,  and 
give  their  sines  and  cosines. 

2.  Name  five  angles  having  their  terminals  symmetric  to  the  horizon- 
tal with  reference  to  the  terminal  of  45°,  and  give  their  sines  and  cosines. 

3.  In  Ex.  2,  replace  the  word  "  horizontal "  by  "  vertical,"  and  answer. 

4.  In  Exs.  1,  2,  and  3,  replace  45°  by  30°,  and  answer. 

5.  In  Exs.  1,  2,  and  3,  replace  45°  by  60°,  and  answer. 

6.  Verify  the  following  tables : 


Angle 

0° 

30° 

45° 

60° 

90° 

120° 

135° 

150° 

180° 

Sine 

0 

i 

\/2 
2 

2 

1 

2 

2 

4 

0 

Cosine 

1 

V3 
2 

V2 
2 

i 

0 

-* 

-V2 
2 

-V3 

1" 

—1 

Angle 

0° 

-30° 

-45° 

-60° 

-90° 

-120° 

-135 

-150° 

-180° 

Sine 

0 

-4 

-V2 
2 

-S3 
2 

-1 

-at* 

2 

2 

~\ 

0 

Cosine 

1 

V3 
2 

V2 
2 

h 

0 

-i 

-V2 
2 

-V5 

2 

-1 

And  fill  in  the  following  tables :  > 

Angle 

0° 

210° 

225° 

240° 

270° 

300° 

315° 

330° 

360° 

Sine 

Cosine 

Angle 

0° 

-210° 

-225° 

-240° 

-270° 

-300° 

-315° 

-330° 

-360° 

Sine 

Cosine 

Note.  —  The  student  is  advised  against  attempting  to  memorize  the 
results  of  the  preceding  tables  as  independent  facts.  It  is  better  to  hold 
in  mind  the  diagrams,  and  read  from  these  mental  diagrams  the  numeri- 
cal values  of  the  sines  and  cosines,  attaching  the  proper  sign  according 
to  the  location  of  the  terminal. 


214  PLANE   TRIGONOMETRY.  [§  117 

7.  Determine  from  the  diagrams  the  smallest  angles  (numerically) 
satisfying  the  following  expressions,  together  with  some  two  other  values ; 
also  the  general  values : 

sin-1— ±  sin"1— ;  ±  sin^O  ±  2  sin^;  3  sin"1 0  ±  4  cos"1 0 ; 

±  2  sin-1  (  -  ^?)  ±  3  cos-1^? ;  ±  sin"1  (1)  ±  5  cos"1  (1) ; 

±  sin"1  (J)  ±  2  cos-1  (|) ;  3  sin"1  (  - 1)  -  2  cos"1  (  -  |) ; 

2  sin(-  2^)  ±  3  «*"*(-  ^?);  sin"1  (0)  +  2  sin"1^?  -  3  cos^? ; 

sin-1  0  +  sin"1  —  +  sin"1  —  +  sin-1  -  +  cos-1  0  +  cos"1 2j=  +  cos"1^?, 

ti     i  £  2,  mm 

and  this  last  when  the  sines  and  cosines  are  negative. 

8.  Solve  the  exercises  at  the  end  of  §  75,  changing  the  word  "  sine  " 
to  "  cosine  "  where  possible. 

9.  Verify  the  following  results  (using  diagrams  and  not  the  tables) : 

(a)  sin  30°  cos  60°  +  cos  30°  sin  60°  =  sin  90°  =  cos  0°  =  -  sin  270°. 

(b)  sin  45°  cos  60°  ±  cos  45°  sin  60°  =  ^  (1  ±  V3)- 

(c)  sin  60°  cos  30°  +  cos  60°  sin  30°  =  sin  90°. 

(d)  cos  60°  cos  30°  +  sin  60°  sin  30°  =  cos  30°. 

(e)  cos  60°  cos  30°  -  sin  60°  sin  30°  =  cos  90°. 
(/)  sin  45°  cos  0°  ±  cos  45°  sin  0°  =  cos  45°. 

(g)  cos2  30°  +  sin2  30°  =  1. 
(h)  cos2  45°  +  sin2  45°  =  1. 
(i)  cos2  60°  +  sin2  60°  =  1. 
(J)  cos2 120°  +  sin2 120°  =  1. 
(k)  cos2  30°  -  sin2  30°  =  cos  60°. 
(I)  cos2  45°  -  sin2  45°  =  cos  90°. 
(m)  cos2  60°  -  sin2  60°  =  cos  120°. 


/  \  ™  a      *r\o        A        .11  —  cos  A   m  A    MA        .     /l  +  cos /I 
(n)  If  A  =  60°,  sin—  =  +*V 5 ,  and  cos—  =?  +yj— ■L— 

(0)  Can  you  show  that  sin2 A  +  cos2 A  =  1,  where  A  is  any  angle? 

10.  Determine  from  the  diagrams  the  value  of 

sin  A  +  cos  B, 

and  sin  A  cos  B  ±  cos  A  sin  B, 

when  A,  B  are  any  pair  of  angles  appearing  in  the  tables  on  page  213. 

Do  not  use  the  table  for  values,  but  get  the  values  direct  from  a  diagram. 


§118]  THE   COSINE  FAMILY.  215 

11.  Verify  from  diagrams  the  following  results,  when  A  =  30°,  45°, 
60°,  or  any  angle  of  tables,  page  213 : 

(a)  sin  2  J.  =  2sinvl  cos  4. 

(b)  sin  3  A  =  3  sin  A  —  4  sin8  A. 

(c)  cos 2 A  =  cos2 A  —  sin2 A. 

(d)  cos  2^4  =  1  -2  sin2 A. 

(e)  cos  2  A  =  2  cos2  A  -  1. 

(/)  cos  3  A  =4  cos8  A  —  3  cos  A. 

§  118.    Complementary  Angles. 

Any  pair  of  angles  whose  sum  in  degree  measure  is  90°,  or 
radian  measure,  — ,  are  called  a  complementary  pair.  Each  is 
said  to  be  the  complement  of  the  other. 

Their  form  is  A°,  90°  -A°;  0,^-6. 

To  construct  the  terminal  of  90°  —  A°,  lay  out  from  the 
upright  vertical  an  angle,  in  the  reversed  direction,  equal 
to  A.  The  terminals  of  an  angle  and  its  complement  are, 
therefore,  inclined  to  the  right-hand  horizontal  line  and  the 
upright  vertical  in  opposite  equality. 

Thus,  if  the  upright  vertical  were  made  the  initial  line 
and  abscissa  axis,  and  counter-clockwise  motion  turned  into 
clockwise  motion,  and  vice  versa,  angles  would  become  their 
complementary  angles  by  equal  turns. 

This  is  equivalent  to  changing  ordinates  into  abscissas, 
and  abscissas  into  ordinates,  or 

sines  into  cosines 

and  cosines  into  sines. 

Thus,    the  sine  of  an  angle  is  the  cosine  of  its  complement^ 

and        the  cosine  of  an  angle  is  the  sine  of  its  complement. 

EXERCISES. 

1.  Show  on  a  diagram  that  if  equal  moduli  are  taken  on  the  terminals 
of  an  angle  and  its  complement,  the  ordinate  of  one  modulus  is  the 
abscissa  of  the  other,  and  vice  versa.     Use  each  quadrant. 


216 


PLANE   TRIGONOMETRY. 


[§119 


2.  Show  that  the  sine  (cosine)  of  any  angle  can  always  be  expressed 
in  terms  of  the  sine  or  cosine  of  an  angle  less  than  45°.  Illustrate  by 
diagrams,  with  the  terminal  in  each  quadrant.  Make  up  and  solve  five 
numerical  examples. 

This  is  made  use  of  in  trigonometric  tables ;  columns  headed  "  sine  " 
at  the  top  are  marked  "  cosine  "  at  the  bottom.  The  teacher  may  point 
this  out  in  whatever  five-place  table  is  used,  showing  that  such  tables 
need  go  only  to  45°  direct  and  that  the  sine  (cosine)  of  any  angle  less 
than  360°  can  be  taken  directly  from  the  table,  by  the  aid  of  formulas 
(i),  (ii),  (iii),  page  201.     See  special  arrangement  in  Hussey's  Tables. 

§  119.    The  Complementary  Arc. 

In  the  early  treatment  of  trigonometry,  with  the  unit 
circle,  and  with  arcs  used  instead  of 
angles,  their  numbers,  or  numerical 
measures,  being  the  same,  arcs  like 
AP  and  BP  (Fig.  119)  were  taken 
as  complementary  arcs.  NP,  the 
sine  of  the  arc  BP,  is  equal  to  the 
cosine,  OM,  of  the  arc  AP.  So  also 
the  sine,  MP,  of  the  arc  AP  is  the 
same  in  length  as  ON,  the  cosine  of  the  arc  BP.  The  com- 
plementary arc,  BP,  behaved  toward  the  upright  vertical  just 
as  the  arc  itself  to  the  horizontal.  The  eo-  of  cosine  thus 
indicated  the  sine  of  the  complementary  arc. 


Fig.  119. 


EXERCISES. 

1.  Select  a  terminal  in  each  quadrant,  construct  its  complementary 
terminal,  and  show  from  the  diagram  that 

sin  (90°  -  A)  -  cos  A  ;  cos  (90°  -  A)  =  sin  A, 

using  both  the  ratio  definitions  and  the  line  definitions. 

2.  Select  at  random  two  angles  (involving  minutes  and  seconds)  in 
each  quadrant,  and  find  their  complements. 

3.  Test  by  diagrams  the  relations  sin  (90°  -  A)  =  cos  A,  cos  (90°- A) 
=  sin  ^4,  for  each  of  the  following  angles  in  degree  measure,  taking  each 
both  positive  and  negative :  0;  30;  45;  60;  90;  120;  135;  150;  180; 
210;   225;   240;   270;   300;   315;   330;  360. 


§119]  THE   COSINE   FAMILY.  217 

4.  Show  that  if  sin  A  =  cos  B,  the  terminals  of  A,  B  are  equally 
inclined  to  the  vertical  and  horizontal  respectively,  but  that  A,  B  are  not 
necessarily  complementary.  Find  the  general  formula  connecting  A,  B, 
when  sin  A  =  cos  B,  and  when  cos  A  =  sin  J3,  as  in  Ex.  5. 

5.  Solve  the  equation  sin  2  0  =  cos  3  0  by  sines. 

Solution. 
sin  2  0  =  cos  3  0  =  sin  (*  -  3  0  Y 

Thus  ^  —  3  6  must  be  some  one  of  the  angles  which  have  the  same  sine 
as  2  0.     2 

...  £-30  =  2n7r  +  20;  (1) 

or  !-30=(2n  +  l)7r-20.  (2) 

Remembering  n  is  any  ±  integer,  its  sign,  in  transposing,  is  not  con- 
sidered. 

.-.  by  (1),    50=(2n+i)7r;  (3) 

by  (2),       0=(2n-\)^  (4) 

by  (3),       0=(4n  +  l)^;  (5) 

or  by  (4),       0=(4n-l)|.  (6) 

6.  Locate  the  terminals  for  a  few  of  the  angles  determined  by  (5), 
(6),  of  the  preceding  solution  and  show  that  the  terminals  of  2  0,  3  6  are 
equally  tilted  toward  the  upright  vertical  and  right-hand  horizontal, 
so  that  sin  2  0  =  cos  3  0. 

7.  Show  from  (5)  and  (6),  of  Ex.  5,  that  2  0,  3  0  as  determined  by 
(5)  are  complementary  only  when  n  =  0,  giving  0  =  18°,  and  that  as 
determined  from  (6)  they  are  never  complementary. 

8.  Solve  Ex.  5  by  cosines  instead  of  by  sines. 

9.  Give  the  general  solution  of  sin  A  =  cos  B  by  sines,  and  show  that 
A  +  B  =  (4  n  +  1)90°  or  A  -  B  =  (4  n  +  1)90°.  For  what  value  of  n  do  we 
get  complementary  angles?    For  what  value  would  we  get  A  =  90°  +  B ? 

10.  Show  from  the  results  of  Ex.  9  that  sin  ((4n  +  1)90° +.4)  =cos  A ; 
sin  (90°  +  A)  =  cos  A  ;  cos  ((4  n  +  3)90°  +  A)  =  sin  A  ;  cos  (270°  +  A)  = 
sin  A .     Compare  with  the  formulas  on  page  201. 

11.  Give  the  general  solution  of  sin  A  =  cos  B  by  cosines,  and  deduce 
the  results  in  Ex.  10. 


218  PLANE   TRIGONOMETRY.  [§  120 

12.  Solve  by  sines  or  by  cosines  the  equations : 

—  sin  A  =  cos  B ; 
sin  A  =  sin  B ; 
sin  A  =  —  sin  5 ; 
cos  A  =  cos  5 ; 
cos  ^4  =  —  cos  -S ; 
and  deduce  the  results  of  §  108,  as  in  Ex.  10. 

13.  Solve  by  sines  or  by  cosines  the  following  equations : 

sin  5  A  =  cos  6  A  ; 

sin  7 A  =  —  cos  9  A  ; 
sin  (sA)  =  cos(L4)  ; 
sin  (sA)  =s  —  cos  (tA). 

14.  State  the  general  method  of  solving  equations  like  those  in  Exs. 
12  and  13.     It  is  given  in  Ex.  5. 

§  120.     Calculations  using  the  Cosine,  without  the  Use  of 

Tables. 

From  the  relation  of  definition  connecting  the  three  quan- 
tities, cosine,  abscissa,  modulus,  when  any  two  of  the  quanti- 
ties are  given,  the  third  can  be  calculated: 

(i)    When  modulus  and  cosine  are  given,  abscissa  =  modulus 

times  cosine. 

abscissa 
modulus 
abscissa 


(ii)  When  modulus  and  abscissa  are  given,  cosine  = 
(iii)  When  abscissa  and  cosine  are  given,  modulus  m 


cosine 

EXERCISES. 

Change  the  words  "  ordinate  "  and  "  sine  "  in  the  examples  of  §  76,  to 
"  abscissa  "  and  "  cosine,"  respectively,  and  solve. 

§  121.     Calculations  with  Cosines,  using  the  Tables. 

These  calculations  may  be  made  either  with  the  tables  of 
natural  cosines  or  with  logarithms.  As  the  student  is  now 
calculating  for  calculation's  sake,  merely  to  learn,  he  may 
use  both  processes  with  each  example,  the  one  as  a  check  on 
the  results  of  the  other. 


121]  THE   COSINE   FAMILY.  219 


For  natural  cosines 


abscissa  =  modulus  times  cosine, 

abscissa 

cosine  = — — , 

modulus 

modulus  =^m. 
cosine 


For  log  cosines : 

log  abscissa  =  log  modulus  +  log  cosine, 
log  cosine      =  log  abscissa  —  log  modulus, 
log  modulus  =  log  abscissa  —  log  cosine. 

EXERCISES. 

After  the  manner  of  the  corresponding  exercises  on  the  sine,  §  79,  carry 
out  the  following  calculations,  with  tests  of  accuracy. 

Let  calculated  parts  show  the  same  number  of significant  figures  as  the  data. 
In  one-figured  data  on  lines,  read  angles  to  the  nearest  five  degrees  ;  in  two- 
figured  data  on  lines,  read  angles  to  the  nearest  half-degree ;  in  three-figured 
data  on  lines,  read  angles  to  the  nearest  five  minutes;  in  four-figured 
data,  to  the  nearest  minute ;  in  five-figured  data,  to  the  nearest  second,  etc. 
(See  §  77.) 

A.  Find  the  general  value  of  the  angle  in  degrees  and  in  circular 
measure  in  terms  of  ir: 

1.  Modulus  9,  abscissa  ±  7 ;  modulus  8,  abscissa  ±  3. 

2.  Modulus  34,  abscissa  ±  23 ;  modulus  6.7,  abscissa  ±  3.4. 

3.  Modulus  23.4,  abscissa  ±  15.4 ;  modulus  4.56,  abscissa  ±  2.34. 

4.  Modulus  23.78,  abscissa  ±  17.34 ;  modulus  2.674,  abscissa  ±  1.789. 

5.  Modulus  234.98,  abscissa  ±  189.90  (five-place  table). 

6.  Modulus  453.764,  abscissa  ±  389.031  (six-place  table). 

7.  Construct  a  set  of  examples  similar  to  the  six  preceding  and  con- 
sistent as  representing  measurements. 

B.  Calculate  the  abscissa,  given : 

8.  Modulus  5,  angle  ±  25° ;  modulus  9,  angle  ±  40°. 

9.  Modulus  34,  angle  ±  55°;  modulus  7.8,  angle  ±  67°. 

10.  Modulus  35.7,  angle  ±  43°;  modulus  3.42,  angle  ±  67°  25'. 

11.  Modulus  23.45,  angle  ±  67°  23' ;  modulus  271.8,  angle  ±  45°  32'. 

12.  Modulus  234.67,  angle  ±  34° 45' 56"  (five-place  table). 

13.  Modulus  4578.67,  angle  ±  23° 45'  56.7"  (six-place  table). 


220 


PLANE   TRIGONOMETRY. 


[§122 


14.  Construct  a  set  of  examples  similar  to  Exs.  8-13  preceding  and 
consistent  as  representing  measurements. 

15.  Express  the  angles  of  Exs.  8-13  in  terms  of  7r;  also  in  radian 
measure. 

C.   Calculate  the  modulus,  given : 

16.  Abscissa  7,  angle  15° ;  abscissa  9,  angle  45°. 

17.  Abscissa  34,  angle  23°;  abscissa  7.9,  angle  70°  30'. 

18.  Abscissa  45.7,  angle  34°  50' ;  abscissa  3.89,  angle  59°  5'. 

19.  Abscissa  34.91,  angle  73°  23';  abscissa  6.789,  angle  45°  21'. 

20.  Abscissa  45.831,  angle  56°  56'  56"  (five-place  table). 

21.  Abscissa  2345.67,  angle  34°  45'  56".7  (six-place  table). 

22.  Abscissa  45.69879,  angle  67°23'34".51  (seven-place  table). 

23.  Construct  a  similar  set  of  consistent  examples  representing  meas- 
urements. 


§  122.     Calculations  with  Right-angled  Triangles,  using  the 
Sine  and  Cosine. 

Right-angled  triangles  may  be  solved  by  using  the  sine 
B      alone,  as  already  shown  in  §  81,  or  by 
using  the  cosine  alone,  since  the  acute 
angles  are  complementary. 

c.  side  opposite  the  angle 

oine  = , t= > 

hypothenuse 

p    .      _  side  adjacent  to  the  angle 
hypothenuse 

We  have  thus  for  the  solution  of  right-angled  triangles : 


sin  A  = 
sin  B  — 


USING   SINES. 

a 
I 
b 
h 
a  =  sin  A  •  h 

b  =  sin  B  •  h 

h  = 


sin  A      sin  B 


USING   COSINES. 

cos  A  =  - 
h 

cos£=y 
h 

a  =  cos  B  •  A 
b  =  cos  A  •  h 
h  = 


cos  B     cos  A 


§123] 


THE   COSINE   FAMILY. 


221 


It  is  customary  to  use  a  combination  of  sines  and  cosines,  the  sine  in 
connecting  the  hypothenuse  and  the  side  opposite  an  angle  and  the 
cosine  in  connecting  the  hypothenuse  and  the  side  adjacent  to  the  angle. 
We  may  think  of  opposite  the  angle  as  in  the  angle,  and  adjacent  to  the 
angle  as  on  the  angle.  As  in  is  in  sine  and  the  o  of  on  is  in  cosine,  you 
will  remember  that  the  side  in  the  angle  involves  the  sine  and  hypothe- 
nuse, while  the  side  on  the  angle  brings  in  the  cosine  and  hypothenuse. 
This  may  be  helpful  when  the  triangles  are  tilted  away  from  a  hori- 
zontal (vertical)  position. 

ORAL  EXERCISE. 

Practise  with  the  cosine  and  sine  in  connection  with  the  Conradt  col- 
lection of  right-angled  triangles  on  page  145. 


§  123.     Calculation  Scheme  for  Right-angled  Triangles,  using 
Log  Sines  and  Log  Cosines. 


Given 


Model  Examples. 
A 


f  B  =  35°  12' 
h  =  27.34 


A  =  90°  -  B. 

b  =  sin  B  -  h. 
a  =  cos  B  •  h. 


Fig.  121. 


Find 


A  =54°  48' 
b  =  15.76 
a  =  22.34 


■.  log  b  =  log  sin  B  -f-  log  h, 
\  log  a  =  log  cos  B  +  log  h. 


Solution  Scheme, 
logarithms.  log-check. 

a-  sin  B  =  b'SinA. 

LOGARITHMS. 


(4)    b 

1.1975 

(2)  sin  B 

1.7607 

(1)    h 

1.4368 

(3)  cos£ 

1.9123 

(5)    a 

1.3491 

1.3491 
1.7607 


1.1975 
1.9123 


1.1098  1.1098 


222  PLANE   TRIGONOMETRY.  [§  123 

Order  of  work  is  indicated  by  the  numerals.  First  enter  the 
logs  (1),  (2),  (3).  Then  add  (1),  (2),  for  (4),  and  (1), 
(3),  for  (5).  Then  look  up  the  numbers  corresponding  to 
logs  (4),  (5),  and  enter  the  results  in  the  column  marked 
"Find." 

If  A  takes  the  place  of  B  in  the  scheme,  a,  b  are  inter- 
changed. 

The  order  of  filling  in  the  scheme  will  change  with  the 
data,  but  not  the  scheme  itself. 

EXERCISES. 

Arrange  the  data  and  results  to  be  found,  diagram,  formulas,  log- 
formulas,  and  calculation  scheme  for  the  following  examples.  Give 
scheme  for  other  possible  cases,  omitting  the  case  when  two  sides  are 
given. 

1.  b  8,  B  65°;  b  6,  B  35°;  a  5,  B  55°;  a7,A  25°. 

2.  b  12,  B  46°;  b  3.4,  B  76°  30';  a  2.7,  B  61°;  a  26,  A  26°  30'. 

3.  b  12 A,  B  45°;  b  34.5,  B  56°  50';  a  678,  A  45°  15'. 

4.  b  23.45,  A  34°  45';  b  345.7,  A  45°  53'. 

5.  b  234.56,  B  56°  34'  31". 

6.  h  8,  B  45°;  h  9,  A  15°. 

7.  h  23,  B  34°;  h  4.5,  A  67°  30'. 

8.  h  23.8,  B  23°;  h  765,  ^  34°  25'. 

9.  ft  236.9,  5  56°  21' ;  I  4.798,  A  34°  51'. 

10.  h  234.59,  5  45°  51'  34";  h  34.761,  ^L  48°  48'  48". 

11.  h  234.786,  B  45°  43'  21".3. 

SINE  AND  COSINE  PROBLEMS  IN   PHYSICS. 

(Such  of  these  as  do  not  lie  outside  the  students'  training  in  physics 
may  be  used.) 

1.  A  force  F  acting  at  an  angle  A°  to  &  horizontal  plane  has  what 
vertical  and  horizontal  components  ?  Solve  a  numerical  example,  getting 
the  sums  of  the  horizontal  and  vertical  components  when  three  forces 
act  at  a  point  in  a  plane  in  different  directions  in  the  plane.  Select  the 
directions  so  that  some  of  the  resolved  forces  are  negative. 


§  123]  THE   COSINE   FAMILY.  223 

2.  A  right-angled  triangle  in  a  vertical  plane,  with  its  hypothenuse 
horizontal  and  right  angle  above,  has  a  uniform  chain  covering  the  other 
two  sides.  Find  the  pressure  of  the  chain  on  each  side,  and  also  the 
force  acting  down  each  side,  and  see  if  a  continuous  chain  on  such  in- 
clines would  be  in  "  perpetual  motion." 

3.  A  string  inclined  at  an  angle  A°  to  a  smooth  plane  inclined  B°  to 
the  horizontal  sustains  a  weight,  W,  on  the  plane.  Find  the  pull  on  the 
string  and  the  pressure  on  the  plane,  and  also  in  what  position  the  string 
is  most  effective  in  preventing  the  weight  sliding  down  the  plane. 

4.  A  mule  hitched  to  a  canal-boat  is  effective  only  75%.  What  is 
the  angle  of  the  pulling  rope  with  the  tow-path? 

5.  Three  horses  of  equal  strength,  and  two  mules,  the  one  as  strong 
as  the  other  and  the  two  as  strong  as  the  three  horses,  pull  on  five  ropes 
tied  to  another  rope  which  passes  over  a  pulley  and  is  attached  to  a 
weight,  W.  The  angles  of  the  five  ropes  with  the  single  rope  are  in 
order  for  the  horses  17°,  19°,  21°  on  the  same  side,  and  for  the  mules,  on 
the  other  side,  18°,  20°,  —  all  in  the  same  plane.  How  many  times  the 
pull  of  a  horse  is  the  lifted  weight,  and  will  the  single  rope  maintain  its 
direction  as  the  animals  pull? 

6.  The  resultant  of  two  forces  is  8.  One  of  them  is  5,  and  the  direc- 
tion of  the  other  35°  from  the  resultant.  What  is  the  angle  between 
the  two  forces  ? 

7.  Two  velocities  represented  by  100,  125,  give  a  resultant  velocity 
153  directly  north.  What  equations  connect  the  angles  that  the  com- 
ponents make  with  the  north  and  south  line? 

8.  The  force  816  makes  angles  25°,  35°  with  its  components.  Calcu- 
late the  components. 

9.  A  movable  pulley  rests  on  a  rope  with  one  end  fixed,  the  other 
end  of  the  rope  passing  over  a  smooth  peg  in  a  horizontal  line  with 
the  fixed  end  of  the  rope.  A  weight  is  attached  to  the  movable  pulley. 
Find  the  equation  connecting  the  pull  on  the  rope  and  the  weight. 
Does  the  pull  increase  or  decrease  as  the  weight  rises  ? 

10.  A  sphere  of  weight  W  and  radius  a  rests  in  the  angle  of  two 
smooth  planes  inclined  at  angles  A°,  B°  to  the  horizon.  Find  the  press- 
ure on  each  plane. 

11.  A  man  weighing  200  lbs.  stands  at  the  apex  of  an  ordinary  tri- 
angular roof  (equal  rafters).  What  part  of  his  weight  tends  to  spread 
the  walls  ? 

12.  When  a  roof  like  that  in  Ex.  11  is  covered  uniformly  with  snow, 
what  part  of  the  weight  of  the  snow  tends  to  spread  the  walls  ? 


224  PLANE   TRIGONOMETRY.  [§  124 

§  124.     Application  of  Sines  and  Cosines  to  a  Problem  in  Field- 
surveying  (a  Projection  Problem). 

What  is  meant  by  the  bearing  of  a  course  of  a  survey  has 
already  been  explained  (§  86).  While  the  latitude  of  a 
course  is  its  length  multiplied  by  the 
cosine  of  its  bearing,  its  longitude  is  its 
length  multiplied  by  the  sine  of  its  bear- 
ing, —  bearings  being  taken  from  the  north 
and  south  line.  In  the  diagram,  the  lati- 
tude AL  is  AP  cos  B,  and  the  longitude, 
or  departure,  AM  is  AP  sin  B,  B  being  the  bearing. 

North  latitudes  are  counted  plus,  and  south  latitudes 
minus ;  east  departures,  plus  ;  west,  minus. 

In  a  closed  survey,  carried  out  with  absolute  accuracy,  the 
algebraic  sum  of  the  latitudes  is  zero,  as  is  the  algebraic  sum 
of  the  departures.  In  practice  this  is  never  reached.  The 
difference  between  the  sum  of  the  northings  and  the  sum  of 
the  southings  is  called  the  error  in  latitude;  the  difference 
between  the  sum  of  the  eastings  and  sum  of  the  westings, 
the  error  in  departure.  The  square  root  of  the  sum  of  the 
squares  of  these  two  errors  is  called  the  error  in  closing.  If 
the  survey  is  plotted  on  a  diagram,  the  error  in  closing  is 
the  distance  between  the  beginning  and  end  of  the  plot 
when  expressed  in  the  unit  of  length  in  which  the  lengths 
of  the  courses  are  given. 

The  error  of  closing  is  a  measure  of  the  accuracy  of  the 
field  work.  When  it  lies  within  the  limit  allowable  for  the 
kind  of  land  surveyed,  it  is  customary  to  "balance  the  survey." 
This  consists  in  distributing  the  errors  of  latitude  and  depar- 
ture among  the  courses  in  proportion  to  their  length  as  com- 
pared with  the  length  of  the  border  of  the  survey. 

The  following  calculation  scheme  applies  to  each  course. 
The  scheme  for  the  entire  survey  is  made  by  placing  such 
schemes  one  under  another  in  the  order  of  the  survey.  At 
the  bottom  of  such  a  scheme  will  appear  the  errors  in  lati- 
tude and  departure,  with  the  error  in  closing. 


124] 


THE  COSINE  FAMILY. 


225 


(4)  D 

(2)  sin  B 
(1)  l 

(3)  cos B 

(5)  L 


Logarithms 

Latitude 

Departure 

Balauoe  Errors 

Balanced 

N  + 

S  — 

£  + 

w- 

Lat. 

Dep. 

Lat. 

Dep. 

i 

l 

1 
1 

i 
I 
l 

I . 

Fig.  123. 


Under  Balance  Errors  and  Balanced  Latitude  and  Depar- 
ture, the  proper  sign  is  given,  along  with  the  magnitudes. 

The  numerals  indicate,  as  heretofore,  the  order  of  entry. 
Logs  (1)  and  (2),  added,  give  (4)  ;  (1)  and  (3),  added,  give 
(5).  Then  the  numbers  corresponding  to  (4)  and  (5)  are 
looked  up  and  entered  under  Departures  and  Latitudes,  in 
the  proper  columns. 

LABORATORY  EXERCISE. 

If  apparatus  is  obtainable,  make  a  survey  of  a  few  fields,  and  carry  out 
the  calculations  as  indicated  above.  Get  also  the  area  of  each  survey. 
The  best  laboratory  for  calculation-trigonometry  is  the  field.  A  real 
stump  and  a  real  river  are  much  more  difficult  to  manage  than  paper 
stumps  and  paper  rivers. 

EXERCISES. 

Fill  out  the  scheme  for  the  following  survey,  the  numbers  1,  2,  3,  etc., 
indicating  the  courses  in  order.  Calculate  the  latitudes  and  departures 
of  the  courses,  and  the  errors  of  latitude,  departure,  and  closing.  Bal- 
ance the  survey.     Make  the  plot. 

1.  N.  69°  E.,  437  ft.  3.   S.  27°  W.,  244  ft. 

2.  S.  19  E.,  236  ft.  4.   N.  71°  W.,  324  ft. 

5.   N.  19°  W.,  184  ft. 

Do  the  same  for  the  following  survey : 

1.  S.  89°  55' E.,.  25.42  ch.  4.  N.  86°  50'  W.,  25.58  ch. 

2.  N.  27°40'E.,  34.68  ch.  5.  S.  47°  30'  W.,  1.50  ch. 

3.  N.  19°  10'  W.,  7.40  ch.  6.  8.  77°  45'  W.,  13.60  ch. 

7.  S.  89°  W.,  3.53  ch. 


226 


PLANE   TRIGONOMETRY. 


[§125 


§  125.    Use  of  the  Cosine  as  a  Check  on  Solutions  of  Triangles. 

In  the  triangle  ABO,  if  OD  is  perpendicular  to  AB,  AD 
is  called  the  projection  of  AC  on  AB,  or  the  projection  of  b 
on  c.     DB  is  the  projection  of  a  on  c. 


AD—  b  -cos A. 
DB  =  a  •  cos  B. 

But  AD  +  DB  =  AB,  for  all  positions  of  A,  B,  D,  when 
signs  are  taken  into  account. 

,\  c  =  a  •  cos  B  +  b  •  cos  A     (1), 
and  b  =  a  •  cos  C  +  c  •  cos  A     (2),  by  symmetry ; 

a  =  b  •  cos  C  +  c  •  cos  B     (3),  by  symmetry. 

Since  the  cosine  of  an  angle  between  90°  and  180°  is 
negative,  the  preceding  formulas  hold  for  all  triangles, 
when  signs  are  taken  into  account.     In  Figs.  124,  125,  and 

AD  +  DB  =  AB, 

in  all  cases. 

EXERCISES. 


Solve  the  following  examples,  and  use  (1),  (2),  (3)  above  to  test  the 
correctness  of  the  solutions : 

1.  Given  B,  41°  41';  6,43.21;  c,  49.32. 

2.  Given  ,4,37° 37';  a,  37.37;  6,41.24. 

3.  Given  ,4,135° 24';  c,  45.45;  a,  54.54. 

4.  Given  a,  37.38;  b,  38.39;  c,  40.41. 

5.  Given  B,  137°  37';  6,  50.51;  a,  30.31. 

6.  Construct  some  appropriate  examples  where  the  sides  have  one 
significant  figure;  two;  three;  five;  six;  seven.  Solve.  Determine  in 
each  case  how  close  the  test  should  hold. 


126] 


THE   COSINE   FAMILY. 


227 


§  126.     Use  of  the  Cosine  in  Connection  with  a  Projection 
Proposition  of  Plane  Geometry. 

In  all  books  on  ordinary  geometry  will  be  found  a  proposi- 
tion reading  somewhat  like  this  :  The  square  on  the  side  of  a 

triangle  opposite  an       ,  [  angle  is  equal  to  the  sum  of 

(  obtuse  J  f  Hi    •   •  n  d  ) 

the  squares  on  the  other  two  sides  \  .  ,        by  twice 

I  increased     J 

the  rectangle  contained  by  one  of  these  sides  and  the  pro- 
jection of  the  other  on  it. 

The  trigonometric  statement  for  this  proposition  is  the 
same,  whether  the  angle  is  obtuse  or  acute,  and  is  this: 
The  square  on  the  side  of  a  triangle  is  the  sum  of  the  squares 
of  the  other  two  sides  diminished  by  twice  the  product  of  these 
two  sides  and  the  cosine  of  their  included  angle. 

When  the  angle  is  obtuse,  its  cosine  is  negative,  so  that 
the  trigonometric  and  geometric  statements  agree. 

Proof  :  In  Fig.  127,  CD  is  perpen- 
dicular to  AB,  and  the  lengths  are  as 
indicated. 

a2  =  p2  +  y2 

=  p2  +  (e  —  x)2 

=  jt?2  +  rc2  +  c2_2^ 

=  &2+(?2_2c.&.COS^.  (1) 

For  p2  +  x2  =  b2, 

and  x  =  b  •  cos  A. 


or, 


In  Fig.  128,  taking  signs  into  account,  AD  +  DB  =  AB, 

C 

x  +  y  =  c. 

\P\b 


—  c  —  x\ 


and  no  change  is  made  in  the  proof. 


228  PLANE   TRIGONOMETRY.  [§126 

We  have  thus : 

a2  =  &2  +  c2_2&c.Cos^,  •         (1) 

62  _  c2  +  a2  _  2  ca  .  cos  B,  (2) 

c2  =  «2  +  52  _  2  ab  ■  cos  b  (3) 

Are  these  formulas  adapted  to  logarithmic  calculation  ? 

EXERCISES. 

1.  Given  a  12,  b  27,  C  30°,  find  c  to  two  significant  figures  without 
using  the  tables.  Do  the  same  when  C  has  the  value  45° ;  60° ;  120° ; 
135° ;  150°.        ' 

2.  Given  a  9,  &7,  C35°,  find  c  to  one  significant  figure. 

3.  Given  a  12,  c  13,  £  48°  30',  find  b  to  two  significant  figures. 

4.  Given  b  23.4,  c  54.1,  ^4  56°,  find  a  to  three  significant  figures. 

5.  Determine  to  one  significant  figure  the  cosine  of  the  angles  of  a 
triangle  whose  sides  are  7,  8,  9.  Can  the  angles  be  determined  uniquely 
(§  71)  from  these  cosines?  Why  is  the  case  different  when  the  sine 
of  the  angle  of  a  triangle  is  given  ? 

6.  Show  that  the  diagonals  of  a  parallelogram  are 

a2  +  b2  ±  2  ab  cos  6,  where  6  is  the  angle  between  the  sides  a,  b. 

7.  Determine  the  diagonals  to  the  appropriate  number  of  significant 
figures,  given  a  3,  &5,  035°. 

8.  Show  that  twice  the  length  of  the  median  of  a  triangle  (from  angle 
A)  1S  VA2+c2  +  2  6ccos^l. 

9.  If  a,  b,  c,  d  are  the  sides,  in  order,  of  a  quadrilateral  inscribed  in 
a  circle,  show  that  if  B  lies  between  a,  b, 

(1)  a2  +  b2-2abcosB  =  c*  +  d2  +  2cdcosB. 

(2)  cosB  =  a2  +  b*-c2-d2. 
v  2(ab  +  cd) 


§  127]  THE  COSINE  FAMILY.  229 

(3)  sin2 B  =  (a  +  b  +  c-d)(b+c  +  d~a)(c  +  d  +  a-h)(d  +  a  +  b-c) 
1  '  ±(ab  +  cdy2 

(4)  (ab  +  cd)  sin  B  =  2  V(s  -  a)  (s  -  6)  (s  -  c)  (s  -  rf) , 
where  2s  =  a  +  6  +  c  +  d. 


(5)  Area  of  quadrilateral  =  V(s  —  a)  (s  —  6)  (s  —  c)  (s  —  cQ. 

(6)  Find  the  area  when  the  sides  are  3,  5,  7,  9. 

(7)  Show  that  the  sine  of  the  angle  between  the  diagonals  is 

2  V Q  -  a)  (s  -  b)  Q  -  c)  (s  ^~d) 
ab  +  cd 

10.   Show  that  the  sines  of  the  angles  which  the  median  of  a  triangle 
(from  the  angle  A)  makes  with  sides  c,  b  are 

asinC  asinJ5 


Vb2  +  c2  +  2  be  cos  A     V62  +  c2  +  2  6c  cos  A 

11.   Two  circles  of  radii  a,  b  cut  each  other  at  an  angle  0.    Prove  that 
the  length  of  their  common  chord  is 

2  ab  sin  0 


Va2  +  b2  +  2  ab  cos  0 


12.  The  revolving  crank-arm  R  of  a  stationary  engine  is  connected 
by  a  link-bar  I  to  a  piston  rod,  whose  stroke  S  is  in  a  straight  line  not 
passing  through  the  centre  of  revolution  of  the  crank-arm.  Draw  the 
positions  of  the  crank  at  the  dead  points  and  show  that  a  triangle  will 
be  formed  whose  sides  are  S,  I  —  R,  I  +  R.  Show  that  if  6  is  the  angle 
between  the  positions  of  the  crank-arm  at  the  dead  points, 

52  =  2  (I2  +  R2)-2  (I2  -  r2)  cos  6. 

Show  that  S  —  2  R  when  the  stroke  line  passes  through  the  centre  of 
revolution  of  the  crank-arm. 

§  127.     Determination  of  the  Angles  of  a  Triangle  from  the 
Cosines  of  the  Half  Angles. 

Since  there  is  only  one  angle  less  than  180°  and  positive 
which  has  a  given  cosine,  the  formulas  of  the  preceding  sec- 
tion determine  the  angles  of  a  triangle  uniquely  from  the 
cosine.  However,  they  are  not  suitable  for  use  when  the 
sides  are  expressed  by  numbers  of  more  than  one  or  two 
significant  figures.  The  cosine  of  the  half  angle  is  suitable 
for  logarithmic  computation. 


230  PLANE  TRIGONOMETRY.  [§  127 

In  §  92  it  was  shown  that  in  Fig.  77 
AD  =  8  —  a, 

and  2&  =  Lz2.hc. 


A  _  AD       Is(s-a) 
2      AO     "V      6c 


Similarly,  cos  |  =  ^  ~  &> 


2       *      «6 


What  sign  must  be  given  to  these  radicals,  A,  B,  C  being 
the  angles  of  a  triangle  ? 

When  all  three  angles  are  to  be  calculated,  these  formulas 
may  be  used  to  best  advantage  in  the  forms : 


A 

cos-  = 

*  abc 

a(s  - 

-  a), 

B 

008  _  = 

* abc 

>b(s- 

-*>, 

0 
cos-  = 

*  aha 

c(s  - 

-o), 

corresponding  to  which  we  have  the  following  calculation- 
scheme,  the  numerals  indicating  the  order  of  filling  in  : 

Calculation-scheme. 

a=               (1)  s-c=  (8) 

b=               (2)  log(«-a)  =  (9) 

(3)  log  (•-*)  =  (10) 

2s=               (4)  log  (•-*)=  (11) 

*=              (5)  loga=  (12) 

«-a  =              (6)  log  5=  (13) 

«-&  =              (7)  log<?=  (14) 


§  127]                             TI 

IE   COSINE   FAMILY.                                231 

log  abc= 

(16),  by  adding  (12),  (13),  (14) 

log«  = 

(15) 

abc 

(17),  by  subtracting  (16)  from  (15) 

log  a(s  —  a)  = 

(18),  by  adding  (9),  (12) 

log  6(s  —  £>)  = 

(19),  by  adding  (10),  (13) 

log  c (s  —  (?)  = 

(20),  by  adding  (11),  (14) 

2  log  cos  |  = 

(21),  by  adding  (17),  (18) 

2  log  cos  -  = 

(22),  by  adding  (17),  (19) 

0 
2  log  cos  —  = 

(23),  by  adding  (17),  (20) 

i            -A 

•'•  log  cos -  = 

(24),  by  taking  half  of  (21) 

log  cos  -  = 

(25),  by  taking  half  of  (22) 

log  COS  -  = 

(26),  by  taking  half  of  (23) 

A 

"   2 

(27) 

B 

2 

(28) 

(7 

2 

(29) 

.-.  A= 

(30) 

B= 

(31) 

C= 

(32) 

Test.     A  +  B+  C-- 

=  180°,  approximately.     (33) 

Note.  —  Here,  as  in  §93,  all  logs,  (9)  to  (15)  inclusive,  are  to  be 
looked  up  before  their  manipulation  begins  with  (16)  and  the  following 
numerals. 


232  PLANE  TRIGONOMETRY.  [§  127 

The  test,  A  +  B  +  0  =  180°,  is  subject  to  the  same  limita- 
tions here  as  in  the  sine-calculation.     (See  page  175.) 


EXERCISES. 

1.  Solve  the  following  by  cosines  and  by  sines  and  compare  results : 

(i)  Given  a,  54.2 ;    b,  59.5 ;     c,  67.3. 
(ii)  Given  a,  9.847 ;  b,  8.352 ;  c,  7.283. 

2.  Solve  by  cosines  the  following  and  test  by  the  formulas  of  §  125 : 

a,  543 ;  b,  586  ;  c,  600.     a,  437.2 ;  b,  564.3 ;  c,  498.3. 
Test  the  same  by  a  sin  B  =  b  sin  A . 

3.  Is  there  any  convenient  way  of  determining  whether  the  test  of 
§  125  is  met  with  sufficient  accuracy  ? 

4.  What  is  an  appropriate  logarithmic  test  for  the  accuracy  of  the 
solution  by  cosines?  If  one  uses  the  test  a  sin  B  —  bsinA,  how  close 
ought  it  to  check  ? 

5.  Make  up  and  solve  a  triangle  requiring  the  use  of  a  five-place  table. 
Check  the  same. 


§  128.   The  Reciprocal  Cosine,  or  Secant,  and  Graphs. 

Related  to  the  cosine  as  the  cosecant  to  the  sine  is  another 
function  of  the  angle  called  the  secant. 

Secant  =  — : — ;     cosecant  = 

cosine  sine 

The  origin  of  the  words  "  secant "  and  "  cosecant,"  as  that 
of  the  "cosine,"  is  clear  on  the  unit-circle  of  the  old-time 
trigonometry. 

Draw  a  tangent  at  the  initial  point  A  of  the  unit-circle. 
Prolong  the  terminal  of  the  angle  to  cut  the  tangent.  The 
secant  of  the  angle,  or  arc,  is  the  amount  cut  off  from  the 
terminal  by  the  initial  tangent. 


128] 


THE   COSINE   FAMILY. 


233 


In  each  of  the  following  diagrams  (Figs.  129-132)  OS  is 
the  secant  of  the  angles  of  the  corresponding  terminals. 


Fig.  129. 


Fig.  130. 


Fig.  131. 


Fig.  132. 


For  ^  =  ^ ;   but  both  OA  and  OP  are  unity,  and  OM 
OA     OM  J 


is  the  cosine. 


,-.  0S=  — : — ,  or  the  secant, 
cosine 


By  looking  up  the  line  picture  of  the  cosecant  (§  96) 
it  will  be  observed  that  the  cosecant  is  the  secant  for  the 
complementary  angle,  or  arc,  just  as  the  cosine  is  the  sine  of 
the  complementary  angle,  or  arc.  Thus  the  co-  of  cosecant 
signifies  the  secant  of  the  complement. 

The  sign  of  the  secant  is  the  same  as  that  of  the  cosine. 
The  range  of  values  of  the  secant  is  determined  by  consider- 
ing that  of  the  cosine  and  taking  the  reciprocal  range. 

As  the  angle  runs  from  0°  to  360°,  the  cosine  runs  over 
the  continuum  from  1  to  0  to  —  1  to  0  to  1. 

At  the  same  time  the  secant  runs  over  the  reciprocal  set 
of  values.  It  goes  from  1  to  positive  infinity,  while  the  angle 
passes  from  0  to  90°.  As  the  angle  passes  over  90°,  the  secant 
makes  a  jump  from  positive  infinity  to  negative  infinity. 
As  the  angle  passes  on  from  90°  to  180°,  the  secant  goes 
from  negative  infinity  to  —  1.  As  the  angle  changes  from 
180°  to  270°,  the  secant  changes  from  —  1  back  to  negative 
infinity.  As  the  terminal  passes  over  270°,  there  is  another 
break  in  the  continuity  of  the  secant,  by  a  spring  from 
negative  infinity  to  positive  infinity.  As  the  terminal  moves 
on  from  270°  to  360°,  the  secant  passes  over  the  continuum 
from  positive  infinity  to  1. 


234 


PLANE   TRIGONOMETRY. 


[§128 


r^ 


w 


37T     27T 

2 


AS  -  A 


m 


Fig.  133. 


If  sin  (90°  + ^4)= cos  J., 
then  cosec  (90°  -f  A)  —  sec  A. 
Therefore,  if  we  look  on 
the  sine  and  cosecant  as 
running  through  a  set  of 
values,  the  cosine  and 
secant  will  be  running 
through  the  same  set  of 
values  90°  behind  the  sine 
and  cosecant,  respectively. 
Thus  the  graphs  of  the 
cosine  and  secant  are  the 
same  as  those  of 
the  sine  and  co- 
secant,    respec- 


tively, if  we  pull  these  back  along  the  horizontal 
axis  a  distance  representing  90°.  (Compare  Figs. 
93,  133.) 

The  diagram  shown  in  Fig.  134,  the  circle  being 
the  unit-circle,  illustrates  the  secants  of  all  angles. 
The  distances  OSv  OS2,  OS3,  etc., 
represent  the  secants  for  the  ter- 
minal positions  OPv  OPv  OPs,  etc., 
and  their  opposites.    When  the  ter- 
minal is  almost  the  upright  vertical, 
the  secant  is  almost  infinite,  positive. 
Just  after  the  terminal  has  passed 
the  position  of  the  upright  vertical, 
the  secant  is  on  the  opposite  ter- 
minal, and  is  thus   negative   and   almost   infinite 
Just  what  the  secant  is  when  the  terminal  is  the 
upright  vertical  or  downright  vertical,  it  is  impos- 
sible to  say.     It  is  customary  to  take  it  as  -f-  x> 
for  90   and  as  -  oo  for  270°. 

Illustrate  on  a  single  diagram  the  line-picture  of  the  sine, 
cosine,  secant,  cosecant  of  an  angle.  Give  such  a  diagram 
for  each  quadrant. 


Fig-  134. 


§  131]  THE   COSINE   FAMILY.  235 

§  129.     Line  Picture  of  the  Secants  of  All  Angles. 

The  secants  are  (Fig.  134)  0A1  •••  0St  -  0S2  •••  0SS~>  +  oo, 
then  -oo  •••  OSr_4  •••  0S_B  •••  0S_2  •••  0S_X  •••  OA,  with  repeti- 
tions. 

EXERCISES. 

1.  Change  the  word  "  sine "  to  "  cosine,"  and  the  word  "  cosecant " 
to  "  secant "  in  the  Exercises  of  §  97,  and  solve. 

2.  By  what  tilt  can  the  graphs  of  y  =  cos-1  x  and  y  =  sec-1  x  be 
obtained  at  once  from  those  for  y  =  cos  x  and  y  —  sec  x  ? 

§  130.     Cosine  Waves  and  Cosine  Harmonic  Motion. 

Since  the  graph  of  the  cosine  is  the  same  as  that  of  the 
sine  pulled  back  a  distance  representing  90°,  the  cosine  may 
be  used  in  connection  with  waves,  just  as  the  sine.  (See 
§§  99,  100.) 

EXERCISES. 

1.  If  y  =  r  •  sin  (at  +  a),  and  y—r-  cos  (at  +  ft),  are  two  waves,  how 
far  behind  one  wave  is  the  other  at  the  same  time-instant? 

2.  Draw  a  diagram  illustrating  the  superposition  of  the  waves  of 
Ex.  1. 

For  harmonic  motion  on  the  horizontal  axis  the  cosine 
plays  the  same  role  as  does  the  sine  on  the  vertical  axis. 
(Reread  §  100.) 

§  131.     The  Versed  Sine,  Exsecant  and  Coexsecant. 

When  the  line  definition  was  in  use,  the  balance  of  the 
radius  (unity)  beyond  the  cosine  was  called  the  Versed  Sine. 

Andsti11  versing  =  1  -  c<M- 

The  versed  sine  is  used  in  engineering  field  books,  as  also 

the  exsecant  =  secant  -  1. 

Also  coexsecant  =  cosecant  -  1. 


236  PLANE   TRIGONOMETRY.  [§132 

EXERCISES. 

1.  Take  the  exercise  on  the  coversed  sine  (§  103),  reading  versed  sine 
for  coversed  sine. 

2.  Show  how  to  plot  a  circular  railroad  curve,  using  sines  and  versed 
sines  as  offsets  (coordinates). 

§  132.     Cosine  Graphs. 

EXERCISES. 

1.  Make  graphs  (Groat's  degree-measure  polar  coordinate  paper)  for : 
r  =  a  cos  0  (circle)  ;  r  cos  6  =  a  (straight  line)  ;  r  =  a  cos  2  $ ;  r  cos  2  6  =  a ; 
r  =  a  cos  3  $ ;  r  cos  3  6  =  a.  See  if  the  number  of  loops  is  affected  by 
n  being  even  or  odd  in  r  =  a  sin  nd  and  in  r  =  a  cos  nd. 

2.  Make  graphs  for : 

r  =  a  (1  +  cos  6)  ;  r  (1  +  cos  0)  =  a, 
r  =  a  (1  —  cos  0)  ;  r  (1  —  cos  0)  =  a. 

(Cardioids  and  parabolas.) 
Show  that  the  cardioid  can  be  drawn  readily  by  means  of  a  circle. 

3.  Make  a  graph  for 


( x  =  a  (0  -  sin  6)  ) 


\  y  =  a  (1  -  cos  0) 

Show  that  the  resulting  curve  is   a  cycloid  (the  curve  described  by  a 
point  in  the  rim  of  a  circular  wheel  rolled  on  a  horizontal  roadway). 

4.  Make  a  graph  for 

r  =  a  sec  0  ±  b.         (a  >  b ;  a  =  b ;  a  <  6.) 

(Conchoid  of  Nicomedes.) 
Show  that  the  conchoid  can  be  drawn  readily  by  means  of  a  straight 
line. 

5.  Make  a  graph  for 

a  nos  0 1 

Show  it  is  a  circle. 

(Use  rectangular  coordinate  paper.) 

6.  Make  a  graph  for 

=  a  cos  6 1 

(An  ellipse.) 


{x  —  a  cos  6 1 
y  =  a  sin  6  J 


(x  =  a  cos  0 1 
y  =  b  sin  0J 


Show  that  points  on  this  curve  are  readily  located  mechanically  by 
means  of  two  concentric  circles  of  radii  a,  b,  prolonging  ordinates  of 
the  one  to  meet  abscissas  of  the  other. 


CHAPTER  VII. 


THE   SINE  AND   THE  COSINE  IN   UNION. 


§  133.     Relation  of  the  Sine  of  Any  Angle  to  its  Cosine. 

In  Fig.  135,  OM2  +  MP  =  OP2.     This  is  true  for  all  posi- 
tions of  the  terminal,  OP. 

Thus,  for  all  positions  of  OP, 
and  for  each  quadrant : 

fOM\2_L_fMP\2_, 

{opJ+Vop)-1' 

or,      (cos  x)2  +  (sin  x)2  —  1,    (1) 
for  all  values  of  x. 

For  any  special  angle,  0,  this 
is  written  in  the  form : 

cos2  6  + sin2  0  =  1,  (2) 

The  relation  (2)  serves  to  de- 
termine the  sine  of  6  when  the  cosine  is  given,  or  the  cosine 
when  the  sine  is  given. 

(3) 

(4) 


2 

2 

2 

2 

2 

2 

2 

2 

/ 

n                              jf 

Fig.  136. 


cos  6  =  ±  Vl  _  Sin-  9, 


sin0  =  ±v/i  _  cos26. 

Explain  on  a  diagram  the  significance  of  the  double 
sign. 

When  the  sine  (or  cosine)  is  given  in  the  form  of  a 
common  fraction  or  a  simple  decimal  (tenths),  it  is  better  to 
calculate  the  cosine  (or  sine)  from  a  diagram  than  from  the 
formulas  (3),  (4). 

237 


238 


PLANE  TRIGONOMETRY. 


[§133 


For  example,  given  the  sine  =  |,  what  is  the  cosine  ? 

Construct  the  terminals  (§§  66  and  111),  assuming  the 
modulus  as  5  and  the  ordinate  as  4  (Fig.  136);  calculate 
the  remaining  side.  Here  it  is  ±  3.  The  cosine  is  thus  ±  f . 
Similarly,  for  sine  —  f,  the  adjoining  diagram  (Fig.  137), 
where  the  cosine  is  ±f. 


V> 


\/. 


t 


^  —  m        ^•"•l*. 

-L**                ^ 

/                          \ 

z                         s 

/                -               ^ 

r                                                        -\ 

7                                                             I 

t                                                             \ 

^6^-              %- 

2\ 

lf\ 

V            ■*■                 -/                V                      A        1 

%   =&        7           \        3    f 

v           Zfr           ]S    -       j. 

\      /X-        3\      ' 

\  -/               v  z 

±z                   V 

^v                ^z 

Fig.  136. 


Fig.  137. 


EXERCISES. 


1.  Determine  by  constructions  the  cosines  and  secants  when  the  sines 
are  the  following,  taking  each  fraction  both  positively  and  negatively : 

A>  tV,  &  B,  H.  0-2,  0.4,  0.75. 

2.  Determine  by  constructions  the  sines  and  cosecants  when  the  cosine 
has  the  following  values,  taken  with  both  signs : 

&,  ih  lb  ti  *i  0.3,  0.1,  0.8. 

3.  Determine  by  constructions  the  secants  and  cosines  when  the  cose- 
cants have  the  following  values,  taken  with  both  signs  : 

II  H,  W-,  -W-,  Hi  Hi 


See  the  table  of  triangles,  page  145,  for  a  large  number  of 
integers  which  may  be  selected  as  the  sides  of  right-angled 
triangles. 

When  the  sine  (cosine)  is  given  in  the  form  of  a  decimal 
of  several  figures,  it  is  better  to  get  the  cosine  (sine)  from 
the  formulas  (3),  (4),  than  from  a  diagram. 


133]         THE   SINE   AND   THE   COSINE  IN   UNION.  239 


EXERCISES. 


1.  Find  the  cosines  to  two  decimal  places  when  the  sine  has  the 
values  (+  or  -):  023)  a36>  a27>  0>21. 

2.  Find  the  sines  to  two  decimal  places  when  the  cosine  has  the 
values  (+  or  -):  a67>  Q^  Q  45>  0QL 


It  will  be  recalled  that  when  z  is  small 


z 


Vl  —  z  =  1  —  -,  approximately  (§  42). 


,.  Vl-^  =  l-|- 


sin2  6 


,\  cos  0  =  1 — ,  approximately,  when  6  is  small. 


EXERCISES. 

1.  Given  sin  10"  =  0.000,048,481,368  ....    Show  that 

sin2 10"  =  0.000,000,002,350,4  ...  and  that 
cos  10"  =  0.999,999,998,825. 

2.  Assuming  (as  will  be  found  true  later)  that  for  the  number  of 
decimal  places  given  above  in  sin  10",  sin  1"  is  ^  of  the  sine  of  10",  and 
sin  n"  is  —  times  the  sine  10",  when  n  <  10  and  positive,  calculate  the 
cosines  of  1",  2",  3", ...  9". 

3.  Eliminate  6  from  x  —  a  cos  0,y  =  a  sin  6.  What  is  the  correspond- 
ing graph  ? 

4.  Eliminate  6  from  x  —  a  cos  6,  y  =  b  sin  6.     What  is  the  graph  ? 

5.  Eliminate  6  from  \  x  =  a  (°  ~  sm  *)  1  .    What  graph  is  this? 

[  y  =  a  (1  —  cos  0)  J  °    r 

6.  Eliminate  t  from   \  y  ~  *  ^      A  !■  •     This  is  the  path  of  a 

[  x  =  vt  COS0  J  r 

projectile.  If  you  know  the  laws  of  falling  bodies  under  gravity,  prove 
that  the  above  equations  are  true  for  a  projectile  started  with  initial 
velocity  v  at  an  angle  6  with  the  horizontal  line. 


240  PLANE  TRIGONOMETRY.  [§  134 

§  134.     Use  of  the  Relation,  (cosine  jt)2  +  (sine  *)2  =  1,  in  trans- 
forming a  Trigonometric  Expression. 

Model  Examples. 


(1)  To  show  that  J} — ^4  =  cosec  A  (1  -  cos  4). 

*  1  +  cos  J. 

Multiplying  the  quantity  under  the  radical,  numerator, 
and  denominator  by  1  —  cos  A,  and  then  writing  in  place 
of  the  new  denominator,   1  —  cos2  J.,  its  value,  sin2  A,  the 

radical  becomes  — \ — - — ,  or,  cosec  .4(1  —  cos^L). 

sini 

(2)  To  show  that  cos4  A  -  sin4  A  +  1  =  2  cos2  A. 
cos4  A  —  sin4  A  =  (cos2  A  +  sin2  A)  (cos2  A  —  sin2  A) 

=  cos2  A  —  sin2  A  —  cos2  A  —  (1  —  cos2  A) 

=  2cos2J.-l. 
From  this  the  transformation  follows  at  once. 
In  working  such  examples,  we  may 

(i)     Transform  the  first  member  into  the  second ;  or 
(ii)    Transform  the  second  member  into  the  first ;  or 
(iii)  Transform  the  first  and  second  members  into  some 
third  expression. 

EXERCISES. 

(Following  from  cos2  A  +  sin2  A  =  1.) 

1.  cos2 ,4  -  sin2 ,4  =  2  cos2 .4  -  1. 

2.  cos2  A  -  sin2  A  =  1-2  sin2  A. 

3.  sec2  A  +  cosec2  ^4  =  sec2  A  cosec2  A 

4.  sec2  A  +  cosec2  A  =  (sin  A  sec  A  +  cos  A  cosec  A)2, 

5.  sin  A  sec  A  +  cos  A  cosec  A  =  sec  A  cosec  A. 

6.  2(1  -cos^)-(l  -cos^4)2  =  sin2^4. 

7    1  -.  sin  A  _  1  +  sin2  A  —  2  sin  A 
1  +  sin  A  cos2  A 

8.  cosec2  A  —  1  =  cos2  A  cosec2  A. 

9.  (sin  J.  —  cos  A)2  =  1  —  2  sin  ^4  cos  ^4. 


§134]        THE  SINE   AND   THE   COSINE  IN   UNION.  241 

10.  sin8  A  +  cos8  A  =  (sin  A  +  cos  A)  (1  —  sin  A  cos  A). 

11.  sin8  A  —  cos8  J.  =  (sin  A  —  cos>4)(l  +  sin  A  cos  .4). 

12.  sin4  A  +  cos4^4  =  1-2  sin2  A  cos2  A. 

13.  sin4  A  —  cos4  A  =  (sin  A  +  cos  A  )  (sin  .4  —  cos  A). 

14.  sin6  J.  +  cos6  A  =  1-3  sin2  A  cos2  A. 

15.  sin6  A  —  cosG  A  =  (sin2  A  —  cos2  A)  (1  —  sin2 ^4  cos2 .4). 

16.  sinM  -  cosM  =  (2  sin2  A  -  1)  (1  -  sin2  A  +  sin4  A), 

17.  sinM  -  cos8  4  =  (sin2 .4  -  cos2 4)  (1  -  2  sin2  A  cos2  A). 

18.  =  sin  A  cos  A. 

cos  A  cosec  A  +  sin  A  sec  ^4 

19        sin^     +l  +  cos^2cosec^ 
1  -f  cos  ^4         sin  A 

and  cos  J     +     cosA         2seeA 

1  —  sin  A      1  -f  sin  ^4 
2Q      1  + sin2 ,4  sec2;!    =  ^ ^  ^ ^ 
1  +  cos2  A  cosec2  .4 

21.  1  ~  sin  ^  =1  +  2  sin  A  sec2  .4  (sin  .4  -  1). 
1  +  sin  A 

22.  sin  A  sec  A  +  sin  B  sec  B      =sin  ^  gec  ^  gin  B  gec  ^ 
cos  ^4  cosec  A  +  cosB  cosec  jB 

23.  sin  ^4  (1  +  sin  yl  sec  A)  +  cos  ^4  (1  +  cos  ^4  cosec  ^4)  =  sec  A  -f 
cosec  vl . 

24.  (sin2  J.  -  cos2.4)2  =  1-4  cos2  A  +  4  cos4  4. 

25.  sin  A  cos  yl  (sin  ^4  sec  A  +  cos  J  cosec  A)  =  1. 

26    sin,4  sec,4  +  sec,4  -  1  =  (1  +  sin^)sec^. 
sin  A  sec  vl  —  sec  A  +  1 


27.  /l-sin^  -  Sec  ^  (1  -  sin  A). 
\  1  +  sin  A 

28.  cosec  4   +  cosec  A      =2sec2A. 
cosec  ^4  —  1  cosec  ^4  +  1 

or*  *  cosec  -4 

29.  cos  ^4  = 


cos  A  cosec  ^4  +  sin  A  sec  ^4 


30.   Ifsin_f4=msin^=       ±m      <andc0Bii=.^ 

cos  4      n  Vm2  +  n2  Vm2  +  n2 


242  PLANE   TRIGONOMETRY.  [§  134 

31.   If  a,  b,  are  any  real  quantities,  angles  A  can  be  found  such  that 

sin  A  =  — — — — — ,  cos  A  = 


Va2  +  b2  Va2  +  b2 

~0    (I  +  cos2  A)2 +(1- cos2  A)2      w      .  .    ,        „  .. 
32'   (1  +  cobM)'  -  (1  -  cobM)«=  ^C0SM  +  SGC  A)' 

33.   (1  +  sin  J.  -  cos^4)2  +  (1  -  sin^4-|-coSi4)2  =  4  (1- sin  A  cos  ^4). 


«4    !  —  sin  yl         cos  A 


cos  .4         1  +*  sin  v4 

oe  1  —  sin^4  cos  A  sin2  A  —  cos2yl        •     A 

35.    • =sinyl. 

cos  A  (sec  A  —  cosec  A)  sm8  A  +  cos3  ^4 

36.  sec2y4  cosec2  A  —  sec2  A  —  2  cos2  A  =  (sin4  ^4  +  cos4  A)  cosec2  A. 

37.  3  (sin  ,4  +  cos  A)  -  2  (sin8  A  +  cos3  ^4)  =  (sin  A  +  cos^4)3. 

38.  sin6  A  +  sin4  ^4  cos2 4  —  sin2  A  cos4  A  —  cos6  A  =  sin2  A  —  cos2 A. 

39.  sin2 4  -  cos2  5  =  sin2£  -  cos2  A. 

40    cos  ^4  +  cos  B     sin  A  —  sin  1?  _  ^ 
sin  J.  +  sin  B     cos  ^4  —  cos  B 

41.  cos2  ^4  +  cos4  J.  cosec2  A  =  cos2  A  cosec2  A. 

42.  sec  ^4  cos3  A  (sec2  ^4  —  1)  =  cosec  A  sin3  ^4 . 

43.  sec2  A  +  cosec2  ^4  =  sin2  A  sec2  ^4  4-  cos2  A  cosec2  A  +2. 

44.  sin2  5  sec2  £=  sec2  B  -  1. 

45.  If  cos  D  =  ^^  and  cos  (90°  -  D)  =  £2*£ 

sinJS  v  y      sin£ 

cos2  A  +  cos2  B  +  cos2  C  =  1. 

46.  If  sin2  ^4  cosec2  B  +  cos2  ^4  cos2  C  =  1, 

then  sin  C  sin  Z?  cos  A  =  sin  ^4  cos  B. 

47.  (1  —  cosec  A  +  cos  ^4  cosec  A)  (1  +  sec  A  +  sin  .4  sec  ^4)  =  2. 

AQ    cos  A  cosec  ^4  —  sin  A  sec  ^4      „„„„„  ,<       „nn  A 

48.   =  cosec  A  —  sec  -4. 

cos  A  +  sin  ^4 

49.  2  (1  -  cos  .4)  +  cos2  A  =  1  +  (1  -  cos  A)2. 

5q    1  —  sin  A  sec  A  _  cos  A  cosec  ^4  —  1 
1  +  sin  A  sec  A      cos  A  cosec  A  +  1 


51.  f     _  1      „,+_  __!_, \oo*A8&*A=1-eo*Aan'A 

1  —  sin2  ^4/ 


k  sec2  A  -  cos2  yl     cosec2  A  -  sin2  ^4  7  2  +  cos2  A  sin2  J. 


§  135]         THE   SINE   AND  THE   COSINE  IN   UNION.  243 

§  135.     Solution  of  Equations  in  Sine  and  Cosine. 
Process :   Reduce  the  equation  to  a  single  function  by 

sin  x  =  Vl  —  cos2  x  or  cos  x  =  Vl  —  sin2  x. 

Example  :  sin  x  +  cos  x  =  1.  (1) 

.*.  sin  x  —  1  =  Vl  —  sin2  x.  (2) 

.\  (sin  x  -  l)2  =  1  -  sin2  x.  (3) 

.-.  2  sin2  x  —  2  sin  x  =  0.  (4) 

•••  sin  x  (sin  x  —  1)  =  0. "  (5) 

.-.  sin  x  =  0 ;  (6) 

or,  sin  x  =  1.  (7) 

By  (6)  z  =  W7r,  or  n  •  180°. 

By  (7)  *=(4n  +  l)f,  or  (4ra  +  l)90°. 

EXERCISES. 

For  Exs.  1-16  give  in  every  case  the  general  solution  in  radian  measure 
(in  terms  of  ir)  and  in  degree  measure,  unless  the  tables  are  necessary 
to  get  the  solution.     Then  give  the  solution  only  in  degree  measure. 

1.  a/3  sin  x  —  cos  x  =  1.  9.  a  sin2  x  +  b  cos  x  +  c  —  0. 

2.  cos2  a;  =  sin2  x.  10.  a  cos2  x  +  6  sin  x  +  c  =  0. 

3.  2  sin  a:  +  cos  x  =  1.378.  11.  cos2  x  -  sin  z  -  J  =  0. 

4.  8  sin  x  =  4  +  cos  x.  12.  2  V3  cos2  x  =  sin  x. 

5.  cos2  Z  =  0.037  sin  x.  13.  2  sin2  x  +  3  cos  a;  =  0. 

6.  sin2  a:  =  0.051  cos  a\  14.  cos2  0  +  cos  $  =  0.673. 

7.  3  sin2  x  -  2  cos  x  -  \  =  0.  15.  sin2  0  -  2  cos  $  +  ,  =  0. 

8.  5  cos2  a:  -  3  sin  x  +  \  =  0.  16.  sec  <£  +  0.34  cosec  </>  =  2.7. 

17.  A  line  was  measured  by  three  different  men.  One  man  reported 
the  length  as  1  mile ;  another,  as  5280  feet ;  the  third,  as  5280.0  feet. 
Were  the  reports  indentical,  so  far  as  care  in  measurement  is  concerned  ? 
Does  it  mean  the  same  to  say  a  race-track  is  a  mile  long  as  it  does  to  say 
that  the  distance  between  two  towns  is  a  mile  ?  Does  it  mean  the  same 
to  say  a  line  is  a  foot  long  as  to  say  it  is  12  inches,  or  12.0  inches,  or  12.00 
inches?  What  effect  on  the  sun's  distance  has  a  change  of  one-hundredth 
of  a  second  in  the  parallax-angle  (angle  subtended  at  the  sun  by  the  earth's 
radius)  ? 


244 


PLANE   TRIGONOMETRY. 


[§136 


§  136.     The  Addition- Subtraction  Formulas  for  Sines,  Cosines. 


sin  (A  +  B)  =  sin  A  cos  B  +  cos  A  sin  B ; 
sin  (A  -  B)  =  sin  A  cos  B  -  cos  A  sin  Z? ; 
cos  (^1  +  -B)  =  cos  ^1  cos  J8  -  sin  ^1  sin  B ; 
cos  (^t  -  B)  =  cos  ^4  cos  B  +  sin  ^4  sin  B. 


(i) 

(2) 
(3) 
(4) 


It  is  necessary  to  prove  only  (1),  (3),  since  they  include 
(2),  (4),  when  A,  B  are  allowed  the  double  sign. 

Now  (1)  is  nothing  but  the  trigonometric  expression  of 
the  geometric  fact  that  if  the  sides  of  a  certain  right-angled 
triangle,  drawn  as  presently  to  be  indicated,  are  projected  on 
any  vertical  line,  the  projection  of  the  hypothenuse  is  the 
algebraic  sum  of  the  projections  of  the  other  two  sides. 
Similarly  (3)  represents  a  projection  of  the  same  triangle  on 
a  horizontal  line.  In  each  case  the  line  on  which  the  triangle 
is  projected  is,  of  course,  in  the  plane  of  the  triangle. 

When  a  line  OP  (Fig.  138),  counted  positive  along  the 
terminal  of  an  angle  #,  and  directed  by  that  angle,  is  pro- 
jected on  a  vertical  line,  its  projec- 
tionis  tfP.sin*. 

If  OP,  as  directed  by  the  angle  x,  is 
for  any  reason  negative,  its  projec- 
tion is  the  opposite  of  the  preceding 
case,  or  —  OP  sin  x, 

or,  OP-  sin  0  +  180°), 

where  the  negative  sign  is  taken  up  by 
the  sine  and  OP  is  now  merely  a  length. 
/.  (a)  Thus,  when  a  length  I,  whose  direction  is  that  of 
the  terminal  of  an  angle  x,  is  projected  on  the  vertical, 

projection  =  I  •  sin  x. 

(/3)  When  a  length  Z,  whose  direction  is  that  of  the 
opposite  terminal  of  the  angle  x,  is  projected  on  the  vertical, 

projection  =  I  •  sin  (a;  +  180°), 

or  I  times  the  sine  of  the  directing  angle  of  the  opposite  terminal. 


136] 


THE   SINE   AND  THE   COSINE  IN   UNION. 


245 


Suppose  that  a,  Fig.  139,  is  the  terminal  of  the  angle  A, 
laid  out  from  the  right-hand  horizontal  line,  and  that  b  is 
the  terminal  of  J5,  laid  out  from 
a  as  a  new  initial  line.  Then 
b  is  also  the  terminal  of  A-\-B, 
as  laid  out  from  the  right-hand 
horizontal  line.  When  a  is  the 
new  initial,  along  it  is  positive; 
opposite,  negative ;  while  also 
A  +  90°  is  a  positive  direction,  and 
A  +  270°  is  a  negative  direction. 

The  triangle  OQP,  whose  projec- 
tions give  (1),  (3),  is  in  all  cases 
drawn  as  follows: 

From  any  point  P,  in  the  terminal  5,  of  A  +  B,  drop  a 
perpendicular  PQ  on  a. 

No  matter  what  the  size  or  sign  of  A,  B,  the  sides  of  the 
triangle  OPQ  can  only  take  four  dif- 
ferent sorts  of  locations  as  follows  : 
OQ  may  be  on  the  terminal  of  J., 
or  on  the  opposite  terminal. 

QP  may  be  directed  by  the 
angle  ^4  +  90°,  as  in  Fig.  139,  or 
by  A  -f-  270°,  as  in  a  diagram  like 
Fig.  140. 

Let  OQ=  T,  or  t,  according  as  it 
lies  on  A's  terminal  or  its  opposite. 
Let  QP  =  Z7,  or  w,  according  as  it  is  directed  by  A  4-  90°, 
or  by  A  +  270°. 

Let  OP  =  r,  which  is  all  the  time  directed  by  A  +  B. 
Thus,  no  matter  what  the  size  or  sign  of  ^4,  B,  there  can 
be  only  four  cases  for  the  triangle  OQP: 

(i)    Sides  r,  T,  U,  (iii)  Sides  r,  fi,  U, 

(ii)  Sides  r,  T,  u,  (iv)   Sides  r,  t,  u. 

Now,  assuming  that  the  projection  of  the  hypothenuse  of 
a  right-angled  triangle  is  the  algebraic  sum  of  the  projec- 


Fig.  140. 


246 


PLANE   TRIGONOMETRY. 


[§136 


tions  of  the  other  two  sides  (§  86),  and  denoting  by  Tv  the 
vertical  projection  of  T,  we  have  for  the  four  possible  cases : 
r9  = T%  +  Uv,  case  (i)  ; 
rv  =  Tv  +  u„  case  (ii)  ; 
rv  =  tv  +  C/;,  case  (iii)  ; 
rv  =  tv  +  uv,  case  (iv). 
By  (a),  p.  244,    2*  =  T  •  sin  A, 
and  U9  =  lf.  sin  (4  +  90°)  =  U  •  cos  A 

By  (£),  p,  244,    tv=t  -sin 4, 
and  w„  =  w  •  sin  (J.  -f  90°)  =  u  cos  J., 

for,  in  the  last  two  cases,  the  directing  angle  of  the  opposite 
of  an  opposite  is  the  angle  itself. 

We  thus  have  the  four  possible  cases : 

r  •  sin  (04  +  B)  =  T  •  sin  A  +  U-  cos  A,  case  (i)  ; 
r  •  sin  ( A  +  B)  =  T  •  sin  A  +  w  •  cos  J.,  case  (ii)  ;    (i?) 
r  •  sin  ( A  +  5)  =  £  •  sin  A  +  £7  •  cos  ^4,  case  (iii)  ; 
r  •  sin  (^4  +  B)  =  t  •  sin  A  +  u  *  cos  ii,  case  (iv). 

Since  now  a  is  the  initial  line  for  the 
angle  B, 

T=r 


cosB 
t  =  r  •  cos  B 
U=  r  •  sin  B 
u  =  r  -  sin  B  > 


,  in  the  appropri- 
ate diagram  ad- 
joining. 


Substituting  these 
values  in  the  equations 
(.27),  we  have,  after  di- 
viding by  r,  the  single 
equivalent  of  them  all: 


sin  (J.  +  B)  =  sin  A  cos  B+ cos  A  sin  B,  (1) 


Fig.  141. 


Formula   (3)    is   now  readily  proven, 
for  when  the  projection  is  made  on  a  hori- 
zontal line,  sin  A  above  becomes  cos  A,  and  sin  (A  -f  90°) 


§  137]         THE   SINE   AND   THE   COSINE   IN   UNION.  247 

becomes  cos  (yl+  90°),  or  —  sin  A,  and  sin  04.  +  B)  becomes 
cos  (A  +  B).     No  other  change  is  made. 

.-.  cos  ( A  +  B)  =  cos  A  cos  B  —  sin  A  sin  B.  (3) 

No  special  proofs  of  (2),  (4)  are  needed,  for  they  are,  as 
already  mentioned,  covered  by  (1),  (3).  When  B  is  nega- 
tive, it  is  added  to  A  by  laying  it  out  clockwise  from  A's 
terminal,  and  no  change  whatever  is  made  in  the  preceding 
proofs. 

§  137.    Special  Cases  of  the  Addition  Theorem. 

The  addition -subtraction  theorems  are  so  important  that 
the  student  should  follow  through  the  proofs  in  some  special 
cases.     For  instance,  for  case  (i),  Fig.  142: 

rv  =  %  +  Uv ; 
or,  OlPl^OlQl  +  Q.P,; 

or,  r  •  sin  (A  +  &)=T-  sin  A  +  U  •  sin  (A  +  90°)  ; 

or,  r  •  sin  (A  +  B)  =  r  •  cos  B  sin  A  +  r  •  sin  B  cos  A. 

.*.  sin  (A  +  B)  =  sin  A  cos  2?  +  cos  A  sin  i?.  (1) 

Similarly,  for  cosines : 

r„  =  Th  +  Uk;                                                  *- 
or,         OP2=OQ,  +  Q2P2.  *£. 

The  student  must  note  care-  /   [\^_ 

fully  that  here,  as  in  all  cases,            /  ^£\ 

the  addition  of  the  projections  is     j^a  '  ' 

algebraic,    OP2  is  shorter  numeri-   o  -^  £2 

cally  than   OQ2,  but   the  added  FlG-  142- 

(J2P2  is  negative.     If  A,  i?,  (7  are  any  three  points  on  a 

straight   line,  then,   always,  no  matter  what  the  order  of 

points,  AB  =  AC+CB. 

The  student,  up  to  this  point  in  his  career,  has  probably  had  this  pre- 
sented to  his  mind  but  seldom.  The  teacher  will  find  it  advisable  here 
to  draw  a  variety  of  diagrams,  showing  that  the  above  statement  is 
always  true  in  algebraic  addition.  When  the  student  has  once  seen 
that  AB  —AC  +  CB,  for  all  possible  arrangements,  the  significance  of 
the  generality  of  the  sin  (J.  +  B)  and  cos  (^4  +  B)  formulas  is  clear. 


248 


PLANE   TRIGONOMETRY. 


[§137 


or, 


In  Fig.  142,       OP2=OQ2  +  Q2P2, 
or,         r  •  cos  (A  +  B^  =  T-  cos^L  +  U-  cos  (A  +  90°), 

r  •  cos  (^4  4-  B)  =  r  •  cos  i?  •  cos  J.  —  r  •  sin  i?  •  sin  A. 
,\  cos  ( J.  +  i?)  =  cos  A  cos  i?  —  sin  A  sin  Z?.  (3) 

Case  (ii),  Fig.  143. 

Tv  =  .Zv  -|-  Wv, 

or,       0^=  0^+  QXPX 

(algebraic  addition). 

Here,  since  a   is   the   initial 

line  for  5,  QP  is  a  negative  line. 

It  is  directed  by  the  angle  A  -f 

270°.     Its  opposite   is   directed 

by  the   angle  .4  +  90°.     Thus, 

by  (/3)  of  the  preceding  section, 

the  projection  of  QP  on  the  vertical  is  QP  sin  (A  +  90°), 

or  u  -  cos  A. 

Thus, 

01P1  =  r -sin  (A +  B~);    0^=  T  -  sin  A;    Q1P1  =  u>  cos  A. 

.\  r  •  sin  (A  -f  i?)  =  I7*  sin  ^4  +  u  •  cos  J.,  as  before. 

But  T=  r  •  cos  B   and   w  =  r  •  sin  i?. 

.*.  sin  (A  +  B)  —  sin  .4.  cos  B  +  cos  J.  sin  B  (1),  as  before. 

Similarly,  in  Fig.  143, 

rh=Th  +  uh, 

or,  OP2  =  OQ2+  Q2P2  (algebraic,  same  as  arithmetic), 

or, 

r  •  cos  (A  +  B)  =  T-  cos  ^4  +  u  •  cos  (^4  +  90°), 

or,  cos  ( J.  +  B)  =  cos  ^4  cos  B  —  sin  J.  sin  B.  (3) 

The  student  may  now  make  diagrams  illustrating  cases  (iii)  and  (iv) 
(bottom  of  page  245),  and  carry  through  the  proofs,  noting  how  many 
different  orders  the  points  Ov  Pv  Qv  and  Ov  Q2,  P2,  may  take,  and  noting 


§  137]         THE   SINE   AND   THE   COSINE  IN  UNION.  249 

that  the  addition  of  segments  is  always  algebraic,  the  algebraic  addition 
occasionally  agreeing  with  the  arithmetic.     See  Ex.  4,  p.  344. 

EXERCISE  IN  PROJECTIONS. 

Show  that  if  a  straight  line  is  drawn  anywhere  in  the  coordinate 
plane  and  a  perpendicular  from  the  origin  is  let  fall  on  it,  making  an 
angle  6  with  the  x-axis,  then  x  cos  0  +  y  sin  9  =  p,  when  p  is  the  length 
of  the  perpendicular,  (x,  y)  being  any  point  on  the  first  line. 

(If  P  is  (x,  y),  T  the  foot  of  the  perpendicular,  and  MP  the  ordinate 
of  P,  project  the  broken  line  OMPTO  on  OT.) 

EXERCISES. 

1.  Given  sin  A  =$,  cos  B  =  T\.  Find,  in  the  form  of  a  common 
fraction,  the  values  of  &in(A  +  B),  sin (4—  B),  cos  (^4  +B),  cos(A-B). 

2.  Solve  Ex.  1  when  cos  A  =  ^,  cos  B  =  f  f. 

3.  Find  the  sines  and  cosines,  cosecants  and  secants,  of  the  following 
angles  in  the  form  of  radicals  and  also  to  two  decimal  places,  without 
using  the  tables :  ±  75°,  ±  15°,  ±  105°,  ±  165°.  What  other  angles 
have  the  same  sine  as  any  one  of  these  angles?  The  same  cosine? 
The  same  sine  and  cosine? 

Answer  in  radicals :    sin  15°  = ;    cosec  15°  =  VQ  +  V2. 

4 

cos  15°  =  V^t^;       sec  15°  =  V6  -  V2. 
4 

sin  75°  =  cos  15°;  cos  75°  =  sin  15°. 

4.  Take,  from  the  tables,  the  sines  and  cosines  of  two  angles,  and 
from  these  calculate  the  sines  and  cosines  of  their  sum  and  difference, 
and  compare  the  results  with  those  of  the  table. 

5.  Apply  the  addition-subtraction  formulas  to  the  angles  90°  ±A> 
180°  ±  A,  270°  ±  A,  360° ±  A.    Compare  the  results  with  those  of  §  108. 

Prove  the  following : 

6.  sin  04  +  B)  sin  (A  -  B)  =  sin2  A  -  sin2  B  =  cos2  B  -  cos2  A. 

7.  cos  (A  -  B)  cos  (A  +  B)  =  cos2  A  -  sin2  B  =  cos2  B  -  sin2  A. 

8.  cos(450-^4)cos(450-JB)-sin(450-^)sin(45°-JB)=sin(^+£). 
(Do  not  expand  the  expressions  in  the  first  member  here  nor  in 
many  of  the  examples  following.) 

9.  sin  (45°  +  A  )  cos  (45°  -  B)  +  cos  (45°  +  A  )  sin  (45°  -  B)  =  cos  (A  -  B) . 
10.  sin  (»  +  1)  A  sin (n  -  1)  A  +  cos  (n  +  1)  A  cos  (n  —  1)  A  =  cos  2  A. 


250  PLANE  TRIGONOMETRY.  [§  137 

11.  sin  (w  +  1)  A  sin  (n  -f  2)  A  +  cos  (n  +  1)  A  cos  (n  -f  2)  v4  =  cos  ^1 . 

12.  cos  a  cos  (y  —  ct)  —  sin  a  sin  (y  —  a)  =  cos  y. 

13.  cos  (a  +  /?)  cos  y  -  cos  (/?  +  y)  cos  a  =  sin  /?  sin  (y  —  a). 

14.  sin  (a  +  /?)  cos  a  —  cos  (a  +  /?)  sin  a  =  sin  /?. 

15.  cos  (a  +  /?)  cos  (a  -  /?)  +  sin  (a  +  /?)  sin  (a  —  /3)  =  cos  2  /?. 

16.  sin  (n  —  1)  a  cos  (n  +  1)  a  +  cos  (n  —  1)  a  sin  (n  +  1)  a  =  sin  2  na. 

17.  sin  (135°  -  0)  +  cos  (135°  +  6)  =  0. 

18.  sin  105°  +  cos  105°  =  cos  45°. 

19.  sin  75°  -  sin  15°  =  cos  105°  -f  cos  15°. 

20.  cos  ,4  +  cos  (120°  -  A)  +  cos  (120°  +  A)  =  0. 


21.  By  setting  sin  (A  +  B  +  C)  =  sin  (A  +  B  +  C),  and  applying  the 
addition  formulas  in  succession,  show  that  sin  (A  +  B  +  C)  =  sin  A  cos  5 
cos  C  +  smB  cos  C  cos  ^4  +  sin  C  cos  ^4  cos  B  —  sin  A  sin  5  sin  C. 

22.  From  Ex.  21,  show  that  if  A  +  B  +  C  =  180°  or  (2  n  +  1)  180°, 
sin  A  sin  5  sin  C  =  sin  ^4  cos  B  cos  C  +  sin  B  cos  C  cos  A  +  sin  C  cos  ^4  cos  5. 

23.  Find,  similarly,  cos  (A  +  B  +  C),  and  give  the  results,  when  here, 
or  in  Ex.  22,  one  or  more  of  the  quantities  A,  B,  C,  are  negative. 

24.  sin  (A  +  B)  +  sin  {A  -  B)  =  2  sin  ^  cos  £. 

25.  sin  (^  +  JB)  -  sin  (^4  -  B)  =  2  sin  B  cos  A 

26.  cos  (A  +  J5)  +  cos  (^4  -  5)  =  2  cos  A  cos  5. 

27.  cos  (A  -  B)  -  cos  (^4  +  B)  =  2  sin  ^4  sin  B. 

28.  V2  sin  (A  -  45°)  =  sin  A  -  cos  A. 

29.  V2  sin  (^4  +  45°)  =  sin  A  +  cos  A. 

30.  sin  (45°  ±  A)  =  cos  (45°  T  -4). 

31.  V2  cos  (.4  +  45°)  =  cos  A  -  smA. 

32.  V2  cos  (J.  -  45°)  =  cos  A  +  sin  ^4 . 

33.  sin  (0  —  <£)  cos  <£  +  cos  (0  —  <f>)  sin  <f>  =  sin  0. 

34.  cos  (0  +  <£)  cos  0  +  sin  (0  +  <£)  sin  0  =  cos  <£. 

35.  2  sin  (a  +  j\  cos  (fi-?\  =  cos  (a  -  /?)  +  sin  (a  +  /?). 

36.  2  sin  (|  -  a\  cos  (|  +  ^\  =  cos  (a  -  ft)  -  sin  (a  +  /3). 

37.  cos(<*  +  /?)  +  sin(e*  -  fS)  =  2sin  (|  +  a\  cos  (f  +  £). 


§  138]         THE   SINE   AND   THE   COSINE   IN  UNION.  251 

38.  cos  (a  +  j3)  -  sin  («-/?)=  2  sin  (|  -  a)  cos  (^  -  /?). 

39.  cos  (n  —  1)  A  •  cos  A  —  sin  (n  —  1)  A  sin  ^4  =  cos  nA. 

40.  If  ,4  +  B  +  C  =  90°,  show  that 

sin2  A  +  sin2£  +  sin2  C  =  1  -  2 sin 4  sin B  sin  C. 

41.  sin  6  sec  0  +  sin  <f>  sec  <f>  =  sin  (0  -f-  <£)  sec  0  sec  <£. 

42.  sin  0  sec  0  —  sin  <£  sec  <f>  =  sin  (0  —  <£)  sec  $  sec  <£. 

43.  cos  $  cosec  6  +  sin  <£  sec  <f>  =  cos  (0  —  <£)  cosec  0  sec  <£. 

44.  cos  6  cosec  0  —  sin  <f>sec<f>  =  cos  (0  +  <j>)  cosec  0  sec  <f>. 

45  sinflsecfl  +  sinfseccfr  =  gin    Q  +  ^  cogec   ^  _  +) 
sin  0  sec  0  —  sin  <£  sec  <£ 

46  sinflsecflsin<frsec<fr  +  l  =  cog  ,$  _  ,  }  gec  (^  +  ,  v 
1  —  sin  0  sec  0  sin  </>  sec  </> 

47  sin  0  sec  0  +  cos  <fr  cosec  <f>  =  cQg  ^  _  #)  gec  ^  +  +) 
cos  <£  cosec  <£  —  sin  0  sec  0 

48  cos  0  cosec  0  +  cos  <j>  cosec  <fr  =  _  gin  ^  +  #)  cogec  (fl  _  #)< 
cos  0  cosec  6  —  cos  <j>  cosec  <£ 

49  1  +  cos  0  cosec  0  sin  <ft  sec  <fr  =  gin  ^  +  +)  gec  (tf  +  +) 

cos  0  cosec  0  —  sin  <£  sec  <£ 

50  1  -  cos  0  cosec  0  sin  <ft  sec  <fr  =  gin   ^  _   . }  gec   Q  _   . } 

cos  0  cosec  0  +  sin  <f>  sec  <f> 

§  138.     The  Addition-Subtraction  Formulas  in  Inverse  Notation. 

If     sin  ( A  +  B)  =  sin  A  cos  B  +  cos  A  sin  i?,  (1) 

then  A  +  B  —  sin-1  (sin  A  cos  i?  +  cos  J.  sin  B).      (2) 

If  sin  A  —  x  and  sin  B  =  y, 

then,  cos  .A  =  ±  VI  —  x2  and  cos  B  =  ±  Vl  —  y2. 

,\  A  =  sin-1  x  —  cos-1  (  ±  VI  —  rr2), 
j5  =  sin-1  y  =  cos-1  (  ±  VI  —  #2). 
Thus  (2)  is  equivalent  to 

sin"1  x  +  sin-1  y  =  sin-1  (  ±  a?  VI  -|/2  ±  y  VI  -  as2) .         (3) 
Similarly,  when  .A  and  B  are  determined  by  sines, 
sin  (A  —  i?)  =  sin  A  cos  i?  —  cos  A  sin  5 
is  equivalent,  in  inverse  notation,  to 


252  PLANE   TRIGONOMETRY.  [§138 


sin-1  x  -  sin-1  y  =  sin-1  (±x  Vl  -  y2  =F  y  >/l  -  as2), 
while         cos  ( A  +  B)  =  cos  ^4  cos  B  —  sin  J.  sin  5 
is  equivalent  to 


sin-1  x  +  sin-1 2/  =  cos-1  (±  Vl  -  as2  VI  -  2/2  -  xy), 
and  cos  ( A  —  B)  —  cos  A  cos  i?  +  sin  A  sin  i? 

is  equivalent  to 


sin-1  x  -  sin-1  y  =  cos-1  (±  VI  -x2  VT-y2  +  a?y). 

iV.5.  What  has  been  said  in  §  73  concerning  the  multi- 
plicity of  values  of  A  when  A  —  sin-1  x  (similar  statements 
holding  when  A  =  cos-1  x)  must  be  borne  in  mind  in  connec- 
tion with  the  results  of  this  section  and  the  examples.  What 
is  meant  by  the  statement  in  Ex.  3,  page  253, 

sin"1 1  +  sin"1  T8y  =  sin"1 £|, 

is,  that  among  the  angles  whose  sine  is  |-|-  is  an  angle  which 
can  be  made  up  by  adding  an  angle  whose  sine  is  f  to  an 
angle  whose  sine  is  y8y.  Similarly,  with  reference  to  the 
statement  A  =  sin-1  x  =  cos-1  Vl  —  x2,  the  proper  sign  is  to  be 
given  the  radical  if  a  special  value  is  to  be  given  to  A.     For 

example,  sin-1  J  is  not  eosW  ^— — J,  if  for  sin"1^  the  special 

value  150°  is  taken.  In  the  exercises  following,  the  radicals 
are  to  be  taken  as  having  both  signs. 

EXERCISES. 

1.  Show  that  if  cos  A  =  x  and  cos  B  =  y,  the  formulas  for  sin  (/I  +  B), 
sin  (A  —  B),  cos  (A  +  B),  cos  (A  —  B)  become,  respectively, 

cos-1  a:  +  cos-1 y  =  sin-1  (#Vl  —  x2  +  xVl  —  y2), 

cos-1  a:  —  cos"1?/  =  sin"1  (yVl  —  x2  —  xVl  —  y2), 

cos-1  a:  +  cos-1?/  =  cos-1  (xy  —  Vl  —  x2  •  Vl  —  y2), 

cos-1  x  —  cos"1  y  =  cos-1  (xy  +  Vl  -  x2  .  Vl  —  y2). 

2.  Give  the  corresponding  formulas  when  sin  A  =  x,  cos  B  =  y,  and 
when  cos  A  =  x  and  sin  B  =  y. 


§138]         THE  SINE  AND   THE   COSINE   IN   UNION.  253 

Prove  the  following,  giving  also  the  general  solution,  as  in  the  model 
solution  (below)  of  Ex.  3. 


Model  Solution  of  Ex.  3. 

Find  the  general  value  of  sin-1 §  -f-  sin-1  T8y.  Let  0  be  the 
principal  angle  of  the  first  quadrant  whose  sine  is  J.  Then 
cos  6  =  | .  Similarly,  if  B  is  the  principal  angle  of  the  first 
quadrant  for  which  sin  <j>  =  ^T,  cos  <j>  =  ^. 

Then  sin"1  f  =  2 mr  +  0,    or  (2w  +  1)tt-0, 

and  sin-1  ^  =  2  mir  +  <£,  or  (2  m  +  1)  tt  —  <f>. 

Then,  sin"1  f  +  sin"1  ^  =  2  sir  +  0  +  <f>,  or  2  *tt  -  (0  +  <£), 
or  (2  8  +  l)7r  +  0-<£,  or  (2  s  + 1)*  -f-  (f>  -  #, 
where  s  is  any  positive  or  negative  integer. 
Now        sin  (2  sir  +  0  +  <£)  =  sin  (0  +  <£), 

sin  (2  87T  -  ((9  +  </>))  =  -  sin  (d  +  <£), 
sin  ((2 1  + 1)  it  +  (0-  £))  =  -  sin  (0  -  </>)  =  sin  (<£  -  0), 
sin  ((2  «  +  l)ir  +  <£  -  0)  =  -  sin  (0  -  0)  =  sin  ((9  -  <£). 
And  sin(0  +  4>)  =  f.lf .  +  |.^  =  JJ, 

sin(^-0)  =  |.TV-f.lf  =  H. 
.•.  sin"1  § -f  sin"1  T^  =  2  m- +  «,  or  (2  r -f- l)7r  —  a, 

where  a  is  some  angle  whose  sine  is  |~£ ;  also  some  angle 
whose  sine  is  —  J-J ;  also  some  angle  whose  sine  is  Jf ;  also 
some  angle  whose  sine  is  —  £§ . 


3.  sin- 1  +  sin- A  =  sin- H.         g_   ^   /T+i+c03.1  JIp: 

4.  sin"1  T55  +  sin-1  ft  =  cos"1 1| |.  s    2  '    2 

5.  cos-1 1  +  cos-4f  =  cos-1  H  10.   C0S-iJg_C0S-i^"+l; 

6.  sin-4  +  cos-4  =  00°.  3  2V3        6 

7.  sin-1 4=  +  cos-1  JL  =  45°.  11.   sin-1  1-  +  sin"1  i|  =  00°. 

v5  vlO  ^,0  J,° 

8.  cos"1  f  -  cos-1  H  =  cos"1  f|.  12.  sin"1  $  -  sin-1  £$  =  sin"1  (?). 


cos-1  a: 


7T 


254 


PLANE   TRIGONOMETRY. 


[§138 


19.  2  sin-1  x  =  cos-1  (1-2  z2). 

20.  2  cos-1  a;  =  cos"1(2x2  -  1). 


13.  sin"1 1  -  cos"1  J?  =  cos-1  (?). 

14.  cos-1^  —  sin-1T8r  =  sin_1(?). 

15.  cos-1  |$  -  sin"1  if  =  cos"1  (?).         21.   cos-1  x  =  2  sin-1  V^T^' 

16.  sin-iff-sin"1!1 ,  =  cos"1  (?). 


22.   cos-1  x  =  2  cos-1 


17.   2  sin-1  x  =  sin-1 2  zVl  -  x'2. 


H± 


18.    2  cos-1  x  =  sin-1 2  zVl  -  a;2. 


23.    sin-^+sin-iA  +  sin-1  !?=?:. 
5  17  85     2 


24.   If  sin-1m  +  sin-1  n  =  ^,  prove  that  mVl  —  n2  +  nVl  —  m2  =  1. 


§  139.   To  find  the  Sine  and  Cosine  of  twice  Any  Angle,  when 
the  Sine  and  Cosine  of  the  Angle  are  known. 

From      sin  ( A  +  B)  =  sin  A  cos  B  +  cos  A  sin  i? 
and  cos  (A  +  i?)  =  cos  J.  cos  i?  —  sin  A  sin  i? 

follow,  by  letting  B  =  A, 

sin  2  ^4  =  2  sin  A  cos  A 


(1) 
cos  2  A  =  cos2  ^  -  sin2  A       (2) 

=  2cos2^-l  (3) 

=  l-2sin2.4  (4) 


A  being  any 
angle. 


That  is,  by  (1),  the  sine  of  any  angle  is  twice  the  sine  of 
half  of  it  into  the  cosine  of  half  of  it,  and  by  (2),  (3),  (4), 
the  cosine  of  any  angle  is  the  cosine  squared  of  the  half  angle 
minus  the  sine  squared  of  the  half  angle,  or  also  twice  the  co- 
sine squared  of  the  half  angle  minus  one,  or  also  one  minus 
twice  the  square  of  the  sine  of  the  half  angle. 


•.  sin  A  =  2  sin 


2VV"2 


cos^  =  cos2^-sin2^» 

cos^  =  2cos2^-l> 

8 


cos^l  =  l-2sin2 


A 


§  139]         THE   SINE   AND   THE   COSINE   IN   UNION.  255 

From  which  follow  the  useful  formulas : 


+  cos  A  - 
-cosA  = 

=  2cos2^, 

=  2  sin2  4' 

8 

staf 

=  ±^l-C08^, 

cos|  = 

=  ±f  +  t*A 

sin^l  = 

=  ±f-t»*A, 

cos4  =  ±V1  +  c^2^ 

EXERCISES. 

1.  Express  the  sine  and  cosine  of  the  following  angles  in  terms  of 
their  half  angles : 

3A;  4A]  5A;  6A;7A',  8A;  nA;  2nA. 

2.  If  sin  6  =  £,  find  sin  2  0  and  cos  2  6,  sin  4  6,  cos  4  $,  sin  8  6,  cos  8  6, 
each  to  two  decimals. 

3.  If  cos  0  =  \,  find  the  values  called  for  in  Ex.  2. 

4.  (sin!±cos^V=l±sin0;  (sin  6  ±  cos  0)2  =  ? 

5.  If  sin  $  =  I  and  cos  <f>  =  \,  find  sin  (0  +  <£)  and  sin  (20+2  <£). 

6.  If  a,  /?  are  positive  acute  angles  and  cos  a  =  \%  and  sin  /?  =  f ,  find 

the  values  of  sin2  ^1^  and  cos2  2-ZJS« 
2  2 

7.  If  a,  /?  are  positive  acute,  and  cos  a  =  f  and  cos/J=$,  find  cos  a~". 
Solve  by  Ex.  8 ;  also  without  using  Ex.  8. 

8.  Prove  that 

(cos  a  +  cos  /3)2  +  (sin  a  -  sin  £)2  =  4  cos2  ^~^j 

(cos a  +  cos /?)2  +  (sin a  +  sin /?)2  =  4 cos2  a~", 
(cos  a  -  cos  /8)2  +  (sin  a  -  sin  /J)2  =  4  sin2  a~P, 


(cos  a  —  cos  /8)2  +  (sin  a  +  sin  /?)2  =  4  sin2 


«+fl. 


256  PLANE  TRIGONOMETRY.  [§  139 

9.  Look  up  the  sine  and  cosine  of  36°  in  the  tables  and  calculate 
sin  72°,  and  compare  the  result  with  the  table  value.  Calculate  cos  72° 
from  the  three  formulas  for  cos  2  A,  and  compare  results  with  each  other 
and  with  that  calculated  from  sin  72°  and  that  of  the  table. 

10.   Cosec2fl=cosece-sec9;   sec2  6=  C0Sef^ "  SeC?fl;    cosec80  = 
2  cosec2  0  -  sec2  0 

sec  4  0  -  sec  2  0  ♦  sec  0  •  cosec  0 
8 
11.   prove      sin2fl  an$Bee$     -^*  ? 

1  +  cos  2  0  1  +  cos  0 

_2    p  cos 2  0     _  cos  0  —  sin  0  ,       cosfl     _  « 

1  +  sin  2  0  ~  cos  0  +  sin  0  '  1  +  sin  0  ~ 
13    P      e      cos  2  0     _  cos  0  +  sin  0  .       cosfl     _  o 

1  —  sin  2  0     cos  0  —  sin  0  '  1  —  sin  0 

_„     -r>  1  +  sec  2  x     0  ^„2       1  +  sec  a;      « 

14.  Prove  — ! =  2  cos2  a; ;  — ■ =  t 

sec  2x  sec  a; 

15.  Prove  sin  A  sec  A  +  cos  4  cosec  A  —  2  cosec  2  ^4 . 

16.  Prove  sin  A  sec  A  —  cos  ^4  cosec  A  —  —  2  cos  2  ^4  cosec  2  ^4. 

17.  Prove  °°^  +  ^  - <*»*-**■*  =  2sin2Asec2A. 

cos  ^1  —  sin  A     cos  -4  -f-  sin  A 

18.  Prove  cosec  2  ^4  (1  +  cos  2  ^4  )  =  cos  A  cosec  ^4. 

19.  prove  -^°     =am(j±*V  sec  ff±^j    -^^=? 

1  T  sin2  0  \4        /  \4        /     l±sin0 

sin  -  +  sin  0 

20.  Prove  -  sin*  ±  sin2  '     =  sin  0  sec  0;    * =? 

1  +  cos0  +  cos20  l  +  cos?  +  cos0 

21.  Prove  cos  (4  +  15°)  cosec  (A  +  15°)  -  sin  (4  - 15°)  sec  {A  -  15°) 
_     4  cos  2  A 

~  1  +  2  sin  2  ^ ' 

22.  Prove  cosec  2  z  (1  —  cos  2  a;)  =  sin  a:  sec  a; ;  cosec  x  (1  —  cos  a)  =  ? 

23.  Prove  sec2  A (1  +  sec  2  A )  =  2  sec  2  A. 

24.  Prove  cosec  ^4  (1  -  cos  2  ^4  )  =  2  sin  ^4 . 

25.  Prove  1  +  cos2  2  0  =  2  (cos4  0  +  sin4  0). 

26.  Prove      cosf^     =  sec  2  A 

cosec2  A  —  2 

27.  Prove  2-^A  =  cos2A;  Ll*gMgf 

secM  aec224 

28.  Prove  ^  +  s;p^-c03f  =  singsecg;    j  ±  sjn  i|  ~  cos^  =  ? 

1  +  sin  0  +  cos  0  2       2      1  +  sin  i  6  +  cos  4  6 


§  139]         THE   SINE   AND   THE   COSINE   IN   UNION.  257 

(A           A  A       A\ 

cos  —  cosec sin  —  sec  —  )  : 
2            2  2        2/' 

cos  2  A  cosec  2  vl  =  ? 

30.  Prove  1~cos2-4  =  sin2 ,4  secM ;  Lni2iil  =  ? 

1  +  cos  2  A  1  +  cosJ. 

31.  Prove      2  Bin  A  gee  il     =  ain2ii. 

1  +  sin2  .4  sec2  ^1 

32.  Prove  ]  ~  s?n^  g^4  =  cos24. 

1  -f-  sin2  ^4  sec2  ^4 

33.  cos2 ^4  (1  +  sin  A  sec  4)2  =  1  +  sin  2  A. 

34.  sin2v4  (cos  ^4  cosec  A  —  l)2  =  1  —  sin  2  4. 

35    /sin  ^4  sec  4  +  IV  _  1  +  sin  2^4 
•    Vsinyl  sec  ^4  —  1/       1—  sin  2  A 

36.  cosM  +  8inM  =  K2-sin2^)» 

cos  ^4  +  sin  ^4 

37.  C08,^-8inM  =  K2  +  sin2^). 

cos  ^4  —  sin  ^4 

38.  cos4  ^4  -  sin4  A  =  cos  2  ^4 ;  cos4  2  A  —  sin4  2A  —  1 

,fl         B  ,   .     .   6  ,      1  +  3cos224 

39.  cos6  A  +  sin6  A  =  — 

4 
40    8in3^4  _  cos  3^4  _  « 
sin  A         cos  A 

41.  cos  3  A  cosec  yl  +  sin  3  A  sec  ^4  a  2  cos  2  ^4  cosec  2  i4. 

42.  sin  4  A  cosec  2  ^4  =  2  cos  2  ^4 . 

43.  sin  5  A  cosec  A  —  cos  5 .4  sec  A  =  4  cos  2  -4. 

44.  sin  5Z  cosec  £  -  cos  —  sec  £  =  2  V3- 

12  12  12        12 

45.  sin  f|  +  tf)  sec  (|  +  *\ -  sin  (?  -  $\  sec(|  -  A  =  2sin20sec20. 

46.  sin  (|  -  fl)  sec  ( j  -  *)  +  cos  (|  -  fl)  cosec  (f  ~  #)  =  2  sec  2  0. 

47.  cos  (^  +  45°)  sec  (A  -  45°)  =  sec  2  ^  (1  -  sin  2  4). 

49.  sin2(45°  +  ^)sec2(450  +  ^)-l  =  sin2^t 
sin2 (45°  +  A)  sec2 (45°  +  A)  +  1 

50.    L^_  +  , 3— r-+, L-+r 1 — ^=? 


1  +  sin2  a;      1  +  cos2  x     1  +  sec2  a;      1  +  cosec2  x 

sin  2  A  —  sin  yi 
1  —  cos^4  +  cos  2^4 


K,        sin  2^4  —  sin^l  o.     A  -^  . 

51. : =  sin  A  sec  J.. 


sln|  = 

-V1- 

cos 
2 

i 

cos=f  = 

=±vl+ 

cos 
2 

A 

258  PLANE   TRIGONOMETRY.  [§140 

§  140.     To  find  the  Sine  and  Cosine  of  the  Half  Angle  when 
the  Sine  and  Cosine  of  the  Whole  Angle  are  known. 

Case  (a).      When  the  cosine  of  the  whole  angle  is  given. 
By  (3),  (4)  of  the  preceding  section 

0) 

(2) 

That  is,  the  sine  of  the  half  of  an  angle  is  the  square  root 
(taken  with  both  sines)  of  one-half  of  the  expression,  one  minus 
the  cosine  of  the  angle,  and  the  cosine  of  the  half  of  an  angle  is 
plus  or  minus  the  square  root  of  one-half  of  the  sum  of  one  plus 
the  cosine  of  the  angle. 

EXERCISE. 

From  the  value  of  the  cosine  of  the  angle  of  a  triangle  in 

a2  =  b2  +c2  -  2  be  cos  A    (§  126) 

A  A 

deduce  by  the  preceding  formulas  the  values  of  the  sin  — ,  cos  —  given  in 

§§  92  and  127.  2  2 

Case  (5).      When  the  sine  of  the  whole  angle  is  given. 
By  (1)  of  §  139. 


Also 


A        A 

2  sin  —  cos  —  =  sin  A. 

A          A 

sin2  —  -f  cos2  —  =  1. 

2              2 

1             1 

(3) 
(4) 

.  sin  ^  +  cos  ^  =  ±  VI  +  sin  A. 

A                 A 

1                 1 

(5) 

sin  ~  —  cos  ^-  =  ±  VI  —  sin  A. 

2             2 

(6) 

.   -«-  ^     ±  VI  +  sin  A  ±  VI  -  sin  A 

(7) 

"*m2-                       2 

cog  A  _  ±  VI  +  sin  A  =f  VI  -  sin  A 

(8) 

§  141]  THE   SINE   AND   THE   COSINE  IN  UNION.  259 

§  141.     The  Significance  of  the  Double  Signs  in  (1),  (2),  (5), 
(6),  (7),  (8)  of  §  140. 

When  any  special  angle  is  taken  as  A,  the  signs  are  not  plus 
and  minus,  but  plus  or  minus,  the  proper  sign  being  deter- 
mined by  the  quadrant  of  the  given  angle.  For  example,  if 
A  =  45°,  the  signs  of  (1),  (2),  (5)  are  all  plus,  and  (6)  is 
minus.  Consequently  the  signs  of  (7),  (8)  are  plus  and 
minus,  plus  and  plus.  Similarly  are  the  signs  settled  for 
(7),  (8)  in  any  other  special  case,  by  first  determining  what 
they  are  in  (5),  (6). 

When  no  special  angle  is  taken  for  A,  but  when  A  is 
assumed  as  given  only  by  its  sine  or  by  its  cosine,  then  the 
reading  of  signs  above  is  plus  and  minus  and  not  plus  or 
minus. 

When  the  cosine  of  A  is  given,  there  are  two  positions  of 
the  terminal,  located  symmetric  to  the  horizontal  (§  106), 
and  the  angles  are  2  n  .  180°  ±  A  (§  106).     Thus  the  half 

angles  are  n  •  180°  ±  ^.     But  the  terminal  for  n  •  180°  +  4 

A 

is   that   of    —  when  n  is  even  and  the  opposite  of  that  of 

A  A 

—when  n  is  odd,  and  the  terminal  of  n  •  180°  —  —  is  similarly 

A 

that  of  —  —  when  n  is  even  and  the  opposite  when  n  is  odd. 

Thus  the  terminals  for  the  half  angles  are  four  lines,  namely, 

A  A 

the  terminals  of  —  and  —  —  and  their  opposites.     The  ter- 

A  A 

minals  of  —  and are  symmetric  to  the  horizontal,  and 

thus  have  opposite  sines  and  the  same  cosine.  Opposite 
terminals  give  opposite  sines  and  opposite  cosines.  Thus 
the  four  terminals  give  only  two  sines  (opposites)  and  two 
cosines  (opposites),  as  shown  in  (1),  (2). 


EXERCISE. 

Illustrate  the  preceding  by  a  diagram. 


260 


PLANE   TRIGONOMETRY. 


[§141 


When  the  sine  of  A  is  given,  there  are  likewise  two  termi- 
nals. They  are  symmetric  to  the  vertical.  The  correspond- 
ing general  angles  are  2  n  •  180°  +  A  and  (2  n  + 1)180°  -  A. 

The  half  angles  are  n .  180°  +  ^andw  180o  +  18Q°~^,  whose 

A 

terminals  are  that  of  —  and  its  opposite,  and  that  of  half  the 

A 

supplement  of  A  and  its  opposite.  This  gives  four  terminals 
for  the  half  angle,  opposite  in  pairs,  and  thus  four  values  of 
the  sine  of  the  half  angle,  opposite  by  pairs,  this  being  also 
the  case  with  the  cosines,  as  shown  in  (7),  (8). 


EXERCISE. 

Illustrate  the  preceding  by  a  diagram. 

Both  the  preceding  cases  are  covered  by  §  53,  where  it 
was  shown  that  for  a  single  position  of  the  terminal  there 
are  two  positions  of  the  terminal  of  the  half  angle. 


§  142.     What  Signs  to  Use  in  the  formulas 

■    i,        A 
■  sin -  + cos- 


±  V 1  +  sin  A, 


sin  — -  —  cos  —  =  ±  Vl  —  sin  A, 
A  A 


(5) 
(6) 


for  a  special  angle,  is  readily  settled  by  a  glance  at  the  sine 
and  cosine  on  the  unit-circle.      The  quadrantal  bisectrices 

mark  the  angles  for  which  there  is 
numerical  equality  in  the  sines  and 
cosines.  For  terminals  between 
—  45°  and  -f  45°,  the  cosine  of  the 
angle  is  positive  and  numerically 
greater  than  the  sine. 

Then  (5)  is  +  and  (6)  is  -. 
The  opposite  is  the  case  for  ter- 
minals between  the  opposites  of  the 
FlQ-  144,  preceding  terminals. 

Thus,  for  terminals  between  those  of  135°  and  —  135°, 
(5)  is  —  and  (6)  is  +  . 


§  142]         THE   SINE   AND   THE   COSINE   IN   UNION.  261 

For  terminals  between  those  of  45°  and  135°,  the  sine  is 
positive  and  numerically  greater  than  the  cosine. 

Then  (5)  is  +  and  so  is  (6). 

The  opposite  is  the  case  between  the  opposite  terminals. 

The  diagram  shown  in  Fig.  144  covers  all  these  cases.  In 
any  special  case  we  need  only  to  locate  the  terminal.  After 
the  signs  of  (5),  (6)  are  settled,  those  of  (7),  (8)  of  §  140 
follow. 

ILLUSTRATIVE  EXAMPLES. 

1.  If  A  =  580°,  what  signs  have  (1),  (2),  etc.,  of  §  140? 

.-.  ^  =  290°,  or  270°  +  20°. 

A 

The  terminal  of  —  lies  between  OR  and  OS  in  Fig.  144,  with  (1)-, 

(2)  + ,  (5)  — ,  (6)  — ,  and  consequently  with  the  order  of  sines  in  (7)  as 
,  and  that  in  (8)  as  —  +  • 

2.  Given   that  the  numerical  value  of  the  sine  and  cosine  at  the 

V2 
quadrantal  bisectrices  is  — ,  or  0.7071,  what  signs  have  (1),  (2),  etc., 

when  sin  A  =  0.3572  (with  cos  A  +),  when  sin  A  =  0.8624,  -  0.4517,  or 
-  0.9653,  with  cos  A  -  ? 

In  the  first  case  the  terminal  lies  between  OS,  OP ;  in  the  second, 
between  OP,  OQ ;  in  the  third,  between  OQ,  OR ;  in  the  fourth,  be- 
tween OR,  OS.  The  corresponding  signs  are  indicated  on  the  diagram. 
(Fig.  144.) 

A 

3.  Where  lies  —  when  the  signs  of  (7)  are ? 

In  this  case  (5)  was  —  and  (6)  was  — .  Thus,  by  Fig.  144,  the  terminal 
lies  between  OR,  OS.    Therefore  the  angle  lies  between 

2rttr-^   and   2n7r-^- 


EXERCISES. 

1.  State  the  signs  for  the  formulas  considered  above  when  A  has  the 
following  values  in  degree  measure  : 

44,  45,  382,  560,  -  44,  -  840,  2040,  -  650,  690,  200. 

2.  Determine  the  signs  for  (5),  (6),  (7),  (8),  when  A  lies  between 
270°  and  315°. 


262  PLANE   TRIGONOMETRY.  [§  143 

3.  The  same,  if  possible,  when  A  lies  between  270°  and  360°. 

4.  What  are  the  signs  in  (7)  when  A  lies  between  450°  and  495°  ? 

5.  Where  must  —  lie  when  the  signs  in  (7)  are  -\ —  ? 

6.  Show  that  if  A  lies  between  585°  and  630c  the  signs  in  (8)  are  —  + . 

7.  Determine  where  —  must  lie  when  the  signs  in  (7)  are  +  +,  — H, 

8.  Determine  where  —  must  lie  when  the  signs  in  (8)  are  +  +,  H — , 


§  143.     The  Double-angle  in  Inverse  Notation. 

From  sin  2  6  =  2  sin  6  cos  0  it  follows,  if  sin  6  —  x  and 


cos  0  =  ±  Vl  —  #2,  that 


sin2  0=±2zVT--z2. 


•.  20  =  sin-1(±2zVl-z2). 


\  2sm-1a?  =  shr1(±  2a?\/l  -sc2). 


That  is,  among  the  angles  whose  sine  is  ±  2  #Vl  —  x2  will  fall 
the  angle  which  is  twice  any  special  angle  whose  sine  is  x. 

For  example, 

Let  sin  6  = 


+  V2 


cos#  = C+or  —  Y 


If  6  is  selected  as  ^,  cos  6  = 


4'  +V2 


Then  2  sin"1  x  =  sin"1  (  ±  2  rr  Vl  -  z2) 

becomes  2sin-1f J  =  sin'1  f  4-1), 

V+V2/ 


7T 


Among  the  angles,  sin"1(-|-l),  is  — ,  which  is  twice  the 
_  2  -j 

angle  — ,  selected  as  a  special  angle  whose  sine  is '•• 

4  ^5  +v^ 


§  144]         THE   SINE  AND   THE   COSINE   IN   UNION.  263 

Similarly,  from  the  relations 

cos  26  =  cos2 6  —  sin2 0, 
cos2  0=l-2sin20, 

cos20  =  2cos2<9-l, 
follow 

2  sin-1  x  =  cos"1  (1  -  2  a?2), 
2  cos"1  x  =  cos"1  (2  x2  -  1). 

These  relations  are  subject  to  the  limitations  to  which 
attention  has  already  been  invited  in  connection  with  the 
multiplicity  of  values  in  the  inverse  notation. 

EXERCISES. 

Show  that  for  special  values  of  the  angles  the  following  relations  are 
true : 

1.  2  sin-1*  =  cos-1  f  9.  2  sin"1^  =  cos"1 1  if. 

2.  2sin-1|  =  cos-1(-2\).  10.  2  cos"1^  =  cos"1  (-#). 

3.  2  sin-1 1  =  sin-1  |f  11.  lbr*§+2  sin"1  H= sin"1©. 

4.  cos"1 1  +  2  sin"1  |  =  120°.  12.  cos"1|  +  2  cos-1T52=cos-1(?). 


5.  cos-1x  =  2sin-1A/^^-  13'  ^-V^cos-^sin-H?). 

2  14.  sin-1^-2sin-11JV  =  sin-1(?). 

6.  cos-1*  =  2cos-1^— 1£.  is.  cos-1T\-2cos-1T51  =  cos-1(?). 

7.  2sin-1^  =  sin-4|f  16.  cos-1 1  -  2  cos"1 1  =  sin"1  (?). 

8.  2sin-1T87=sin-1|f^.  17.  sin-1  (.35)  +  2  cos"1  (.37)  =  ? 


144.  To  find  the  Sine  and  Cosine  of  3  A  in  Terms  of  the 
Sine  and  Cosine  of  A,  respectively,  and  the  Use  of  the 
Results  in  solving  Cubics. 

sin  (3  A)  =  sin  (2  A  +  A)  =  sin  2  A  cos  A  +  cos  2  A  sin  A 
=  2  sin  A  cos  A  cos  A  +  (1  —  2  sin2  A)  sin  A 
=  2  sin  A  (1  -  sin2  ^1)  +  (1  -  2  sin2  A)  sin  4. 

.-.  sin  3  A  =  3  sin  A  -  4  sin3  ^.  (1) 


264  PLANE  TRIGONOMETRY.  [§  144 

A  similar  process,  expressing  the  terms  as  rapidly  as  pos- 
sible in  terms  of  cos  A  instead  of  sin  A  as  above  when  sin  3  A 
is  determined,  gives : 

cos  3  A  =  4  cos3  A  -  3  cos  A~  (2) 

Inasmuch  as  (1),  (2)  have  the  4,  3  terms  interchanged, 
it  may  be  helpful,  in  remembering  them,  to  note  that  each 
term  contains  a  3,  and  that  they  must  both  remain  true  for 
any  special  angle.  Take  for  A  the  angle  whose  sine  is  1  in 
the  sin  3  A  result,  and  for  A,  in  the  cos  3  A  result,  the  angle 
whose  cosine  is  1 ;  namely,  90°  and  0°,  respectively. 

Then  sin  3^4  =  sin  270°  =  -  1, 

so  the  4  must  follow  the  3 ; 

and  cos  3  A  =  cos  0  =  1, 

so  the  3  must  follow  the  4.* 

The  equations  (1),  (2)  show  that  if  sin  3  A,  cos  3  A  are 
given,  the  sine  and  cosine  of  A  are  to  be  obtained  by  solving 
cubics.  Thus  the  ancient  problem  of  "trisecting  an  angle 
by  means  of  the  ruler  and  compass,"  is  the  same  as  the  alge- 
braic problem  of  solving  a  cubic  when  the  second  power  of 
the  unknown  quantity  is  absent,  as  is  the  case  in  (1),  (2). 

Any  cubic  equation,  as 

ax*  +  bx2  +  ex  +  d  =  0, 

can  be  freed  of  the  x2  term  by  setting  x  =  y 

3  a 

becomes,  on  expansion,  an  expression  like 

aff+py  +  q  =  0.  (3) 

Comparing  (3)  with 

4  cos?  4  -  3  cosJ.  -  cos  3  J.  =  0,  (2) 

*  Or  this : 

Next  "  sin  "  comes  "  trin,"  3 
Next  "  co  "  comes  *'  fo,"  4  1 


§  144]         THE  SINE  AND  THE  COSINE   IN  UNION.  265 

cos  A  might  be  y,  if  y  were  less  than  1,  numerically.     We 
may  set  (3)  in  the  form 

a  (ny)z  +  pn2  (ny)  +  qns  =  0,  (4) 

where  n  may  be  so  selected  that  ny  may  be  less  than  1,  and 
thus  the  cos  A. 

When  an  expression  like  (3)  is  zero,  it  remains  zero  when 
multiplied  by  any  number.  Therefore,  when  two  similar 
expressions  are  identical  and  zero,  there  is  proportionality 
among  the  coefficients  of  corresponding  terms.  So  if  (2), 
(4)  are  identical,  and  cos  A  =  ny, 


a      pn2 

4      -  3      - 

qns 
-  cos  3  A 

(5) 

.2           3<*. 
.*.  71  —      - — ; 

4jo 

(6) 

p 

(?) 

(6)  shows  that  n  is  real  only  when  a,  p  are  opposite  in 
sign  ;  (6)  also  shows  that  n  may  have  two  values,  equal  and 
opposite  in  sign.  However,  it  is  necessary  to  take  only  one 
value  of  w,  since  —  n  in  (4)  merely  changes  the  sign  through- 
out and  gives  only  the  same  solutions  as  +  n ;  (7)  shows 
that  no  real  angle,  3  A,  exists  unless  3  qn  <  jt?,  numerically,  or 

9  5%2<  jt?2,  or,  by  (6),  9  q2( -r-^)<p2i  numerically, 

or,  27  (fa  <  4  j93,  numerically.  (8) 

(8)  is  the  test  as  to  whether  a  cubic  of  the  form  of  (3)  can 

be  solved  by  comparison  with  (2),  when  a  is  positive  and  p 

negative,  and  3  A  is  an  ordinary  angle,  —  one  whose  cosine 

is  real. 

Model  Example. 

z3  +  6z2  +  9z+3  =  0.  (!) 

Here  a  =  1,  5  =  6.  .\  to  get  rid  of  the  ofi  term,  set 
x  =  y  —  — -,  or  x  =  y  —  2.     Then  (1)  becomes 

2,3_3y  +  i  =  0.  (2) 


266  PLANE  TRIGONOMETRY.  [§  144 

Here  a  =  1,  p  —  —  3,  q  =  1;  a,  p  are  opposite  in  sign,  with 
(8)  satisfied. 

(6)  gives  n2  =  J.     .\  w  =  J. 

(7)  gives         cos  3  J.  =  —  J. 

.•.3i  =  2m.  180°  ±  120°. 
.v4  mm  •  120°  ±  40°. 

This  gives  only  three  different  cosines  for  A,  —  those  of 
40°,  160°,  280°. 

And  since  ny  =  cos  A  and  n  =  J,  and  x  =  #  —  2,  the  three 
values  of  a;  are  #  =  J  cos  40°  —  2  j  #  =  —  ^  cos  20°  —  2 ; 
#  =  l  cos  80°  —  2.  Looking  up  these  cosines  in  the  table,  we 
have  an  approximate  solution  of  (1),  all  three  values  of  x 
being  real. 

It  is  this  case  —  three  real  roots  —  to  which  the  comparison 
method  pointed  out  above,  for  solving  a  cubic,  is  suited. 
The  student  is  familiar  with  the  fact  that  a  quadratic  has  two 
roots.  So  the  cubic  has  three,  of  which  only  one  may  be 
real,  with  two  imaginary,  or  all  three  may  be  real.  At  least 
one  root  must  be  real. 

To  make  the  solution  of  the  cubic  complete,  we  give  below 
the  so-called  Cardan's  Rule  or  process,  suited  to  the  case  when 
a,  j»,  are  both  positive,  and  to  the  case  where  27  (fa  <  4  p3 
(8)  is  not  satisfied  with  a  positive  and  p  negative. 

Let  ay*+py  +  q  =  Q.  (3) 

Assume         y  =u^-{-vK 

.♦.  y*  =  u  +  3  usv^  -f  3  it M  -f  v, 
or,  yz  =  u  +  v  +  3  wM  (t*»  +  v*), 

or,  j       ys  =  (u  +  v)  -f  3  u*v*  •  y, 

or,  yz  —  3  uM  #  —  (u  +  v)  =  0.  (  XT) 

Dividing  (3)  by  a  and  considering  (3)  and  ( Z7)  as  iden- 
tical, then  i  i       vy 

o  a 

and  _  (M  +  „)  =  1  (£) 


§  144]        THE   SINE  AND  THE  COSINE  IN  UNION.  267 

Cubing  (a)  and  adding  four  times  the  result  to  the  square 
of  (/3),  we  have      2  2     f      4  »3  ,  \ 

<r      27  ad 

Now,  w2-2w  +  v2=(m- v)2,  and,  when  real,  must  be 
positive  or  zero.  The  second  member  of  (7)  is  always  posi- 
tive if  a,  p  are  the  same  in  sign;  but  if  a,  p  are  opposite  in 

sign,  we  must  have  a2      4  »3  .    „ 

2  5ot3'  numencally, 

a1  &  27  a3 

or,  27  aq2  5  4  pB,  numerically.  (9) 

Now,  compare  this  with  (8).     It  is  the  opposite  case. 

When  (9)  holds,  (7)  gives  u  —  v  and  (/3)  gives  u  4-  v. 

Adding  and  subtracting,  etc.,  we  have  u  and  v.  Whence, 
y  from  y=u\  +  vl,  single  values,  when  only  the  ordinary,  or 
real,  cube  roots  of  w,  v  are  used. 

The  student  can  readily  see  that  any  real  number,  like 
w,  v  above,  has  three  cube  roots,  one  real  and  two  imaginary. 

For  suppose  zs  =  u  =  u*l. 

Then  z  =  -v^.^l, 

where  we  suppose  vm  is  the  ordinary  cube  root  of  u.     We 
will  show  that  -\/i  has  three  values. 

Let  83  =  1. 

.-.  «8-l  =  0. 
...  j>-l)(#+t'  +  l)*6. 
,\  8-1  =  0,  or  s2  +  s  +  1  =  0. 
The  first  supposition,  8  —  1  =  0,  gives  8  =  1. 
Solving  the  quadratic  supposition  in  the  usual  way,  we  get 

2       2 

The  three  values  of  8  taken  with  y/%  Vv  give  y  three 
values. 

Summary  for  solution  of  ayz  +  py  +  q=0,  taking  a  positive  : 
(i)    If  p  is  also  positive,  use  Cardan's  process  as  just  given, 
(ii)  If  p  is  negative  and  27a<?2>4jt?3,  numerically,  use 
Cardan's  process. 


268  PLANE  TRIGONOMETRY.  [§144 

(iii)  If  p  is  negative  and  27  aq2  <  4 p%  numerically,  use  the 
trigonometric  method  as  in  the  example  on  page  265. 

(iv)  If  p  is  negative  and  27  a(f  =  4  ps,  use  either  method, 
or  solve  by  the  method  of  equal  roots. 

In  cases  (i),  (ii)  there  will  be  one  real  root  and  two  imagi- 
nary; in  cases  (iii),  (iv),  three  real  roots,  there  being  equal 
roots  in  case  (iv). 

EXERCISES. 

Determine  by  which  method  each  of  the  following  examples,  from 
Ex.  2  to  Ex.  10,  is  best  solved,  and  solve: 

1.  2x8-3x-l  =  0.  6.  x8-3x-2  =  0. 

2.  xs  +  3  x2  -  1  =  0.  7.  xs  -  x  +  6  =  0. 

3.  x8-24x-32  =  0.  8.  x8-  9  x-  28  =  0. 

4.  x8-  7 x  +  5  =  0.  9.  3  x8  -  6  x2-  2  =  0. 

5.  x8  +  4x2+2x-l  =  0.  10.  x8  -  15  x2  -  33  x  +  847  =  0. 

11.  Show  sin  4  0  =  4  sin  $  cos8  0  —  4  cos  0  sin8  0. 

12.  Show  cos  4  0=1-8  cos2  0  +  8  cos4  0. 

13.  Show  cos5  0=16cos50-2Ocos80  +  5cos0. 

14.  Show  cos6  0  =  32cos60-48cos40  +  18cos20-l. 

15.  Show  sin  3  0  =  (2  cos  2  0  +  1)  sin  0. 

16.  Show  cos  3  0  =  (2  cos  2  0  -  1)  cos  0. 

17.  Show  cos3  0  +  3cos0  =  cos8flcosec8fl. 

3  sin  0  -  sin  3  0 

18.  Show  sin  3  0  -  sin  0  =  sin  0  (cos  3  0  +  cos  0)  sec  0. 

19.  Show  sin  3  0  -  cos  3  0  =  (sin  0  +  cos  0)(2  sin  2  0  -  1). 

20.  Show  (1  +  2  cos  0)2(1  -  cos  0)  =  1  -  cos  3  0. 

21.  Show  (sec2  0  +  l)8(3sin0-sin3  0)2 

=  (sec  2  0  -  l)8  (3  cos  0  +  cos  3  0)2. 

22.  Find  expressions  for  sin  5  0  and  sin  6  0  in  terms  of  functions  of  0. 

23.  Show  that  A  =  18°  is  a  particular  solution  of  the  equation 

sin  2  A  =  cos  3  A. 
Express  this  equation  in  terms  of  sin  A  and  cos  A .     Then  in  terms  of 
sin  A,  and  solve,  showing 

sin  A  =  sin  18°  =  +V^~1 


A            1QO      +VlO  +  2V5 
cos  A  =  cos  18°  = -1 . 


§  145]         THE   SINE  AND  THE  COSINE  IN  UNION.  269 

24.  Give  the  general  solution  of  the  equation 

sin  2  A  =  cos  3  A . 

25.  Show  from  the  values  of  sin  18°,  cos  18°,  in  Ex.  13,  that 


sin36O  =  ±Vl0-2Vo. 
4 

cos  36°  =2%!, 
4 

using  sin  2  x  =  2  sin  x  cos  x. 

26.   Show  that        sin  9°  +  cos  9°  =  VI  +  sin  18°, 


sin  9°  -  cos  9°  =  -  Vl  -  sin  18°. 
From  these  equations,  show 


Sin  9o=V3W5-V5  -V5 
4 

c089°  =  V^g  +  V?r71. 
4 

What  are  the  values  of  sin  81°  and  cos  81°? 

§  145.     The  Addition-multiplication  and  Subtraction- 
multiplication  Formulas. 

It  is  frequently  advisable,  either  for  purposes  of  simplifica- 
tion or  to  set  an  expression  in  a  form  suitable  for  logarithmic 
computation,  to  express  the  sum  or  difference  of  two  sines 
(cosines)  in  the  form  of  a  product.     From 

sin  (A  4-  B)  =  sin  A  cos  B  4-  cos  A  sin  i?,  (1) 

sin  (A  —  B)  =  sin  A  cos  B  —  cos  A  sin  B,  (2) 

cos  ( A  4-  &)  =  cos  A  cos  B  —  sin  A  sin  B,  (3) 

cos  QA  —  i?)  =  cos  J.  cos  i?  4-  sin  A  sin  i?,  (4) 

follow  by  addition  (subtraction) : 

sin  ( A  +  B)  4-  sin  (A.  -  ^)  =  2  sin  J.  cos  B,  (5) 

sin  (.4  +  B)  -  sin  (.A  -  B)  =  2  sin  5  cos  A,  (6) 

cos  (A.  -  B)  4-  cos  (A  4-  B)  =  2  cos  A  cos  5,  (7) 

cos  (A  -  B)  -  cos  (i  +  5)  =  2  sin  A.  sin  £.  (8) 


270  PLANE   TRIGONOMETRY.  [§  145 

Here  A  4-  B  and  A  —  B  may  represent  any  two  angles. 
Let  A  +  B  =  x, 

A-B  =  y. 

Then  A  =  X-±M, 

A 


2 
Thus,  (5),  (6),  (7),  (8)  may  be  written  : 

sin  x  +  sin  y  =  2  sin  £±J£  cos  ^=-^>  (5') 

sin  a;  -  sin  2/  =  2  sin  ^=^  cos  ^^'  (60 

cos  2/  +  cos  a?  =  2  cos  ^"t  y  cos  X~V>  (7') 

cos  2/  -  cos  oc  =  2  sin  ^±^  sin  ^i^  C8') 

These  formulas  are  general,  being  true  for  all  values  of  x, 
y.  The  subtractions  in  the  second  members  must  be  carried 
out,  so  far  as  sines  are  concerned,  in  the  order  indicated,  since 
sin  (—  0)  ==  —  sin  6  ;  for  the  cosines,  this  is  a  matter  of  indif- 
ference, since  cos  (—  0)  =  cos  6. 

The  order  in  (7'),  (8;)  is  set  different  from  that  in  (5'), 
(6'),  as  relates  to  x,  y,  in  deference  to  angles  of  the  first 
quadrant,  since  here  the  larger  the  angle,  the  larger  the  sine, 
but  the  larger  the  angle,  the  smaller  the  cosine. 

EXERCISE. 

1.  Give  verbal  statements  for  (5'),  (6'),  (7'),  (8'),  paying  attention 
to  the  order  of  subtractions. 

2.  sin  60°  +  sin  30°  =  2  sin  45°  cos  15°. 

3.  sin  60°  -  sin  30°  =  2  sin  15°  cos  45°. 

4.  sin  25  -  sin  45°  =  -  2  sin  10°  cos  35°. 

5.  cos  60°  +  cos  30°  =  2  cos  45°  cos  15°. 

6.  cos  60°  -  cos  30°  =  -  2  sin  45°  sin  15°. 


§  145]         THE   SINE   AND   THE   COSINE   IN   UNION.  271 

7.  cos  25°  -  cos  55°  =  2  sin  40°  sin  15°. 

8.  sin  115°  +  sin  281°  =  -  2  sin  18°  sin  7°. 

9.  sin  317°  -  sin  151°  =  -  2  sin  36°  cos  7°. 

10.  cos  295°  +  cos  817°  =  2  cos  16°  sin  9°. 

11.  cos  370°  -  cos  280°  =  2  sin  35°  sin  45°. 

12.  Verify  some  of  the  preceding  results  by  using  the  tables. 

13.  Take  a  pair  of  angles  in  each  quadrant  and  apply  (5'),  (6'),  (7'), 
(8'),  letting  (i)  x>y  in  (6'),  (8'),  and  then  (ii)  x<y,  expressing  the 
second  member  in  each  case  in  terms  of  angles  less  than  45°  (if  not  45°). 
Test  by  the  tables. 

14.  Apply  (5'),  (6'),  (7'),  (8')  to  pairs  of  positive  angles  selected  one 
from  one  quadrant  and  the  other  from  another,  in  the  six  possible  ways 
for  (5'),  (7')  and  in  the  twelve  possible  ways  for  (6'),  (8'),  expressing  the 
second  members  in  terms  of  angles  less  than  (or  equal  to)  45°.  Test 
some  of  the  results  by  the  tables. 

15.  Apply  (5'),  (6'),  (7'),  (8')  to  pairs  of  negative  angles  from  the 
same  quadrant  and  to  pairs  from  different  quadrants,  and  test  some  of 
the  results. 

16.  Apply  (5'),  (6'),  (7'),  (8')  to  a  pair  of  angles,  one  positive  and  one 
negative,  making  the  selections  show  variety  of  sign  in  the  second 
members,  after  the  angles  there  are  reduced  to  angles  less  than  45°. 

17.  sin  7  0  -  sin  5  0  =  2  sin  0  cos  6  6. 

18.  sin  0  +  sin  2  $  =  2  sin  ^  cos  ^ 

19.  sin  2  0  -  sin  4  6  =  -  2  sin  6  cos  3  0. 

20.  cos  3  6  +  cos  7  0  =  2  cos  5  6  cos  2  6. 

21.  cos  3  B  -  cos  5  0  =  2  sin  0  sin  4  0. 

22.  cos  7  0  -  cos  5  6  =  -  2  sin  6  sin  6  0. 

23.  cos  2  0  -  cos  0  =  -  2  sin  ?  sin  -—  • 

24.  sin  (45°  +  A)  +  sin  (45°  -  A)  =  V2  cos  A. 

25.  cos  (45°  -  A)  +  cos  (45°  +  A)  =  V2  cos  A. 

26.  sin  (45°  -  A)  -  sin  (45°  +  A)  =  -  V2  sin  A. 

27.  cos  (45°  +  A)  -  cos  (45°  -  A  )  =  -  V2  sin  A. 

28.  cos  (60°  +  A)  +  cos(  60°  -  A)  =  cos  A. 

29.  In  Exs.  24,  25,  26,  replace  45°  by  60°  and  find  second  member. 


31 


272  PLANE  TRIGONOMETRY.  [§145 

30.   Find  second  member  in  Exs.  24,  25,  26,  27,  with  53°  replacing  45°. 
sin  (|  4-  #)  4-  sin  {*2+$\  =  V2  cos  6. 

32    sin  75°  + sin  15°  _  _  cos  30°     _  ^ 
cos 75° -cos  15°         sin  30° 

33.    sin75°-sinl5°=0<577_t 
cos  75°  +  cos  15° 

34     sin  x  +  sin  y  _  «  35    cos  a?  4-  cos  y  _  ^ 

sin  a;  —  sin  y  cos  x  —  cos  y 

36     sin  a:  +  sin  y      «  37     sin  x  —  sin  y  _  ^ 

cos  a;  —  cos  y  cos  a:  —  cos  y 

38.  sin  50°  -  sin  70°  +  sin  10°  =  0. 

39.  sin  10°  +  sin  20°  +  sin  40°  +  sin  50°  =  sin  70°  4-  sin  80°. 

4a   sin7^-sin5^  =  sin^sec^ 
cos  7  A  +  cos  5  A 

41.   sinJ+sm3^  =  sin2^sec2^; 
cos  A  +  cos  3  A 

42      sin7^-smyl  =c034^sec5A 
sin  8  A  —  sin  2  J. 

43.  cos2Z?  +  cos2,4  =  cos  ^  +  j?)  cos  (^  _  B)  cosec  (^  +  5)  cosec  (A-B). 

COS  w  x)  —  COS  —  -id. 

44.  Bin2.B  +  sin24  =  _  gin  ^      ™  CQg  ^j  _  ™  gec  ^  +  ™  CQgec  (^  _m. 
sin25-sin2;l  v  y        v  y        v  '  v  J 

A-    sin  ^4  4-  sin  2  J.  „  „  A  _  M  ^4 

45.   — ^ =  —  cos  —cosec  — - 

cos  2  A  —  cos  A  2  2 

AC    sin  5  A  —  sin  3  ^4        .     A  . 

46.   =smAsecA. 

cos  3^1  +  cos  5  A 

47.  c082A-cos2B  =  -sm(A-B)sec(A-B). 
sin2  J  +  sin25  v  }        v  ' 

48.  cos  (4  +  5)  +  sin  (A  -  B)  =  2  sin  (45°  4-  A)  cos  (45°  +  5). 
4g    cos  3  ^4  —  cos  ^4      cos  2A—  cos  4  ^4  sin  ^4 


sin  3  A  —  sin  ^4      sin  4  vl  —  sin  2  ^4      cos  2  J:  cos  3  A 
50.  sin(4^-25)+sin(45-2^)      sin  ^      B)  gec  ^      ^ 
cos(4yl -25)4- cos(45- 2,4)  v  '        v  ' 

51  cos  ?  d  +  2  cos  5  ^4  +  cos  7 ,4  _  cos  o  ,4  _  sin  2  A  sin  3  ^4 

cos  A  4-  2  cos  3 .4  4-  cos  5  ^4  '  ,  cos  3  ^4 

52  sinfl  +  sinSe  +  sinSe  +  dnT^^^^^, 

cos  0  4-  cos  3  6  4-  cos  5  0  4-  cos  7  0 


§  146]         THE  SINE  AND  THE  COSINE  IN  UNION.  273 

53.  sin  (0  +  <£)-  2sin0  +  sin  (0  -  <f>)  =  ^^^ 
cos  (0  +  <£)  -  2  cos  0  +  cos  (0  -  <f>) 

-A      sin  0  +  2  sin  3  0  +  sin  5  0         .    **  nncfnn  K  a 

54.  .  — — .  =smdt/  cosec  5  0. 

sin  3  0  +  2  sin  o  0  +  sin  7  0 

M     sin  (0  -  <f>)  +  2  sin  0  +  sin  (0  +  d>)         .    ^  0 

55.  - — ±- — 7  (     _   . — „       .    ;„     y,\  =  sm  0  cosec  5. 
sin  (/3  -  <f>)  +  2  sm£  +  sm  (/3  +  <£)  M 

cc    sin  0  —  sin  5  0  +  sin  9  0  —  sin  13  0  A  *  .  a 

56.  - — pi- — j — —  =  cos  4  0  cosec  4  0. 

cos  0  —  cos  5  0  —  cos  9  0  +  cos  13  0 

57.  cos  3  x  +  cos  5  a;  +  cos  7  a:  +  cos  15  x  =  4  cos  4  x  cos  5  a;  cos  6  x. 

58.  sin0  +  sin20  +  sin40  +  sin50  =  4cos-cos  —  sin30. 

59.  Show,  without  using  (5'),  (6'),  (7'),  (8'),  that 

sin  x  -f  sin  y  _  cosy  —  cos  a: 
cos  y  +  cos  x      sin  y  —  sin  x 

cos  0  —  cos  3  0     cos  6  0  —  cos  4  0 


60 


sin  0  —  sin  5  0     sin  8  0  +  sin  2  0 


61  cos(x-f  y+z) +  cos(— x+y  +  z) +cos(x  —  y-f  2)  +  cos  (ar+y— 2)  _  cosy 
sin  (x  +  y  +  z)  +  sin(—  x  +  y  +  z)  —  sin  (a:  —  y  +  z)  +  sin  (a;  +  y  —  z)      sin  y 

g2    cos  0  +  cos  3  0  _     sec  2  0 

cos  0  +  cos  5  0 "~  2  -  sec  2  0 

§  146.  Application  of  the  Addition-multiplication,  Subtraction- 
multiplication  Formulas  to  the  Solution  of  Trigonometric 
Equations  of  a  Special  Type. 


Model  Example. 

Solve 

sin  6  +  sin  7  6  =  sin  4  0. 

(1) 

By  (5), 

2sin40cos30=sin40, 

(2) 

or, 

sin40(2cos30-l)  =  O. 

(3) 

- 

.-.  sin  4  6  =  0, 

(4) 

or, 

2  cos  3  6  -  1  =  0. 

(5) 

By  (4), 

4  0  =  mr.       .*.  0  =  w-. 

By  (5),  cos  3  0  =  |  .         (6) 

A  particular  solution  is  3  0  =  —. 


274  PLANE   TRIGONOMETRY.  [§146 


IT 


,\  the  general  solution  is  .    3  6  =  2  nir  ±  — , 

o 

Thus  the  general  solutions  of  (1)  are 

The  applicability  of  the  process  .indicated  in  this  example 
is  limited  to  the  case  where  a  factor  is  apparent  after  the 
simplification  made  in  (2).     An  example  like 

sin  0  -+  sin  7  6  =  sin  13  0, 
or  like  5  sin  0  +  3  cos  6  =  4, 

or  2  sin  6  +  3  sin  7  0  =  2, 

could  not  be  solved  by  the  process  given  above.  Later,  a 
solution  of  such  expressions  by  the  aid  of  auxiliary  angles 
will  be  given.     (See  §  148  ;  also  §  180.) 


EXERCISES. 

Solve  the  following,  giving  general  values  for  the  angle,  0 : 

1.  cos  0  +  cos  7  6  =  2  cos  4  0.  7.   cos  $  +  cos  2  0  +  cos  3  6  =  0. 

2.  cos  0  +  cos  7  0  =  cos  3  0.  8.   sin  6  +  sin  2  0  +  sin  3  6  =  0. 

3.  sin  0  -  sin  7  0  =  sin  3  0.  9.  cos  0  -  cos  2  0  +  cos  3-0  =  0. 

4.  sin70-sin0  =  cos40.  10.  sin  0  -  sin  2  0  +  sin  3  0  =  0. 

5.  cos  0  -  cos  7  0  =  2  sin  4  0.  11.   cos  0  +  sin  2  0  -  cos  3  0  =  0. 

6.  cos  70-  cos 0  =  sin  3  0.  12.   sin  0  -  cos 2  0  -  sin  3  0  =  0. 

13.   sin  2  0  -  cos  2  0  -  sin  0  +  cos  0  =  0. 

14.  sin  0  +  sin  2  0  =  0.  18.   sin  mO  ±  sin  n0  =  0.     m  =  n. 

15.  sin  0  —  sin  5  0  =  0.  19.   cos  mO  ±  sin  n0  =  0.     m^n. 

16.  cos  20  +  cos  30  =  0.  20.   sin  0  ±  cos  3  0  =  0. 

17.  cos  7  0  —  cos  5  0  =  0.  21.   sin  mO  ±  cos  n0  =  0.     m  =  n. 
22.   sin  (3  0  +  «)  +  sin  (3  0  -  a)  +  sin  («  -  0)  -  sin  (a  -f  0)  =  cos  a. 

23.  cos  n0  =  cos  (n  —  2)0  +  sin  0. 

24.  sintL±i^  =  sill!Lzi^  +  sin0. 


§147]         THE   SINE   AND   THE   COSINE  IN   UNION.  275 

25.  sin2n0  -  sin2(n  -1)0  =  sin20. 

26.  sin  3  0  -  4  sin  0  sin  (0  +  j3)  sin  (0-/3)=  0. 

27.  cos30  +  2cos20  =  O.  31    cos20     sin20  =  1cos2  0 

28.  cos20  +  3cos0  =  O.  "  sin^      cos^        sin2^ 
29    cos_0_sin_0  =  2                           32.  cos  30  -  cos50  =  sin  0. 

'  sin0     cos0  33.  cos  5  0  +  cos  3  0  =  V2  cos  4  0. 

30.  ^  +  ^  +  ^=0.       34.  sin60  +  sin40  =  2cos0. 
cos  0     cos  2  0     cos  3  0 

35.  sin  (m  +  1)0  +  sin  (m  -  1)0  =  cos  0. 

36.  cos  m0  —  cos  (m  —  2)0  =  sin  0. 

37.  sin  0  +  sin  2  0  +  sin  3  0  =  cos  0  +  cos  2  0  +  cos  3  0. 

38.  sin  2  0  -  cos  2  0  =  cos  0  -  sin  0. 

39.  cos  («  -  0)  =  sin  (a  +  0). 

40.  sin  (0  +  a)  +  cos  (0  +  a)  =  sin  (0  -  a)  +  cos  (0  -  a) . 

41.  cos  3  0  sin8  0  + sin  30  cos8  0  =  0. 

42.  sin  0  +  sin  2  0  +  sin  3  0  +  sin  4  0  =  0. 

43.  cos3  0sin80  +  sin3  0cos80  =  — . 

8 

44.  sin  0(cos  2  0  +  cos  4  0  +  cos  6  0  +  cos  8  0)  =  sin  4  0. 

45.  2  cos  2  0  cos  3  0  -  cos  0  =  0. 

46.  4sin20  +  sin22  0=3. 

47.  2  sin80  +  3  cos20  +  sin  0  =  3. 

48.  cos80  -  cos  0  sin  0  -  sin80  =  1. 

49.  sin80  =  0.2341  sin2  0. 

§  147.    The  Multiplication-addition,  Multiplication-subtraction 

Formulas. 

The  formulas  (5),  (6),  (7),  (8)  of  §  145,  when  read  back- 
ward, may  be  called  the  Multiplication-addition,  Multiplica- 
tion-subtraction formulas. 

sin  A  cos  B  =  \  {sin  (A  +  B)  +  gift  (A  -  B)},  (1) 

sin  B  cos  .4  =  !  {sin  (A+B)-  sin  (-4  -  B)},  (2) 

cos^4cos.B  =  |{cos(^i-  J5)  +  cos  (.4  +  .B)},  (3) 

sin  A  sin  B  =  \  {cos  (-*  -  B)~  cos  (A  +  JB)}.  (4) 


_  fee 

For  example,  <    . 

(si] 


276  PLANE   TRIGONOMETRY.  [§  147 

From  (3),  it  appears  that  if  both  words  of  the  product  are 
cosines,  the  product  may  be  changed  to  a  sum  of  cosines; 
from  (4),  that  if  both  words  of  the  product  are  sines,  the 
product  may  be  changed  to  a  difference  of  cosines. 

cos  40°  cos  60°  =  J  (cos  20°  +  cos  100°), 
sin  40°  sin  60°  =  \  (cos  20°  -  cos  100°). 

In  (1),  (2),  the  form  of  subtraction  in  the  second  member 
assumes  A  numerically  greater  than  B,  so  we  may  say  that 
when  the  words  of  the  product  are  one  sine  and  the  other 
cosine,  then,  if  sine  is  connected  with  the  numerically  larger 
angle,  a  sum  of  sines  is  indicated,  and  if  sine  is  connected 
with  the  smaller  angle,  a  difference  of  sines  is  indicated. 
For  example, 

sin  60°  cos  40°  =  }  (sin  100°  +  sin  20°), 
sin  40°  cos  60°  =  £  (sin  100°  -  sin  20°). 
However, 

sin x  cos y  —  \  J  sin (x  +  y)  +  sin  (x  —  y) }, 

no  matter  what  the  relative  size  of  x  and  y,  when  the  differ- 
ence x  —  y  is  taken  in  the  order  indicated. 
For  example, 

sin  40°  cos  60°  =  J  J  sin  (40°  +  60°)  +  sin  (40°  -  60°)  | 
=  l5sinlOO°  +  sin(-20°)} 
=  i  (sin  100° -sin  20°). 

EXERCISES. 

1.  sin  80°  cos  30°  =  ?  9.  sin0cos<£  =  ? 

2.  sin  30°  cos  80°  =  ?  10.  cos0cos<£  =  ? 

3.  cos 80° cos 30°  =  ?  S$       5$ 

4.  sin  80°  sin  30°  =  ?                           ^  ^T^T^ 
5.^70^69  =  1                            12>  sin3icos5i  =  ? 


6.  sin5  0cos70  =  ? 

7.  cos  5  6  cos  7  0  =  ? 

8.  sin  5  0  sin  7  6  =  1 


2  2 


7.   cos50cos70  =  ?  13.   sin  — cos-  =  ? 

2         2 


§  147]         THE  SINE  AND  THE  COSINE  IN  UNION.  277 

14.  Simplify  2  cos  2  0  cos  0  -  2  sin  4  0  sin  0. 

-.,=    a-      nt      -50       0        .90       30 

15.  Simplify  sin  —  cos  -  —  sin  —  cos  —  • 

16.  Prove  2  sin  (45°  +  A)  sin  (45°  -  A)  =  cos  2  A . 

17.  Prove  2cos(45°  +  ^1) cos (45°  -  A)=  cos  2^4. 

18.  Prove  that  sin  -  sin  —  +  sin  —  sin  — —  =  sin  2  0  sin  5  0. 

19.  Prove  cos  2  0  cos  -  —  cos  3  0  cos  -—  =  sin  5  $  sin  — . 

mm  £ 

11/9        $  7/9        3  0 

20.  Prove  sin  — —  sin  -  +  sin  -—  sin  — —  =  sin  2  0  sin  0. 

4  4  4  4 

21.  Prove  2  sin  2  0  cos  0  +  2  cos  4  0  sin  0  =  sin  5  0  +  sin  0. 

22.  Prove  cos  2  0  cos  0  -  sin  4  0  sin  0  =  cos  3  0  cos  2  0. 

23.  Prove  sin  a  sin  («+ 2  /?)  -sin  /3  sin  (/3  +  2a)  =  sin(a-/?)  sin  («  +  /?)• 

24.  Prove  (sin  3  <f>+  sin  <£)  sin  <£  +  (cos  3  <f>  -  cos  <f>)  cos  <j>  =  0. 

2  sin  (0  -  <f>)  cos  <£  —  sin  (0  —  2  <£)  _  sin  0 
2  sin  (/?  —  <£)  cos  <f>  —  sin  (/?-  2<f>)~  sin  /3* 

26    sin  0  sin  2  0  +  sin  3  0  sin  6  0  +  sin  4  0  sin  13  0  _  sin  9  0 


27. 


sin  0  cos  20  +  sin  3  0  cos  6  0  +  sin  4  0  cos  13  0      cos  9  0 

cos  2  0  cos  3  0  —  cos  2  0  cos  7  0  +  cos  0  cos  10  0     cos  6  0    cos  5  0 


sin  4  0  sin  3  0  -  sin  2  0  sin  5  0  +  sin  4  0  sin  7  0      sin  6  0    sin  5  0 

28.  cos  0  sin  (j3-<f>)  +  cos  /?  sin  (<£  -  0)  +  cos  <£  sin  (0  -  )3)  =  0. 

29.  sin  (/?  -  y)  cos  (a  -  8)  +  sin  (y  —  a)  cos  ((3  —  8)  +  sin  (a  —  /?)  cos 
(y-8)=0. 

30.  cos  (0  -  <£)  cos  (0  +  <f>)=  cos2  0  -  sin2  <£  =  cos2  <£  -  sin2  0. 

31.  versin  {0  +  <£)  versin  (0  -  <f>)  =  (cos  0  —  cos  <f>)2. 

32.  2cos^cos|f  +  coSi|  +  eos||  =  0. 

33.  cos  (36°  -  A)  cos (36°  +  A)  +  cos  (54°  +  A)  cos  (54°  -  4)  =  cos  2  A. 

grin  r        sin -*- 

34.  - 


2  2  sin  <£ 


0  +  <b  0  -  <f>     cos  0  +  cos  <f> 

cos— IpE     cos — —3C  n         ^ 

35    sin  0  •  sin  2  0  +  sin  2  0  sin  5  0  +  sin  3  0  sin  10  0  _  _  sin  7  #  sec  7  0 
cos  0  •  sin  2  0  +  sin  2  0  cos  5  0  -  cos3  0  sin  10  0 

sin  5  0  _  sin  3  0  _  sin  2  0  _  sin  2  0  §  sin  3  0  §  sin  5  0 
cos  5  0     cos  3  0     cos  2  0     cos  2  0    cos  3  0    cos  5  0 


278 


PLANE   TRIGONOMETRY. 


[§148 


§  148.    Use  of  an  Auxiliary,  or  Subsidiary,  Angle  in  Solving 
Trigonometric  Equations  of  a  Certain  Type. 

Model  Example. 
2  sin#  +  3  cos#  =  3.  (1) 

Divide  by  the  square  root  of  the  sum  of  the  squares  of  the 
coefficients  of  sin  x  and  cos  x. 


2      • 

—  sin  x  -f 


V13 


cos#  = 


Since 
and 

we  may  set 


V13 

sin2  y  +  cos2  y 

,2 


V13 
1 


(2) 


2        2V13 


sin  y. 
cos  y 


VI3        13_ 

3        3V13 


(3) 

(4) 


V13        13 

Either  (3)  or  (4),  when  expressed  as  a  decimal,  directly  or 
by  logarithms,  determines  y  in  the  tables.     Thus  (2)  becomes 

3        3V13 


sin;?/ sin  x+cosy  cos  x= 

If  we  determine  z  from  the  tables  by  means  of 

3V13 


cos,-     -    , 

(5)  becomes 

cos 

>(x—y)  =  cos  2; 

or, 

x  -  y  =  2  n  180°  ±  z. 
,\  x  =  2  n  180°  +  y  ±  z 

By  (3), 

2V13 

Zj  (6), 

.-.  y  = 

3VT3 

COS  Z  as  — = 

13 

(5) 


(6) 

(7) 
(8) 
(9) 


(The  student  may  determine  «/,  3  from  the  tables.) 
Then,  by  (9),  x  =  2  n  .  180°  4-  #  ±  z  (as  found),  where  w  is 
any  integer. 


§  148]         THE   SINE  AND   THE   COSINE  IN   UNION.  279 

z  and  y  are  called  auxiliary  angles ;  sometimes  subsidiary 
angles. 

Example  (1)  above  belongs  to  the  general  type. 

a  sin  6  +  b  cos  0  =  c,  (^4) 

whose  solution  is,  similarly, 

a        sinfl+        h        cosfl  =        g 
V«2  +  J2  Va2  +  #>  Va2+~P 

Let  sin  <f>  =  ,  cos  6  =  —  , 

V^+~P  Va2  +  62 

and  cos  a 


Va2  +  b2 
•\  cos  (0  —  <£)  =  cos  «. 

.-.  0  =  2  mr  ±  a -\- (j>. 

Since  no  cosines  are  greater  than  unity,  evidently  a  cannot 
be  determined  if  c2  >  a2  -f-  £>2.     In  this  case  no  solution  exists. 

When  #,  b  are  large,  use  §  180. 

• 

EXERCISES. 

A.  Show  that  (^4)  above  may  be  solved  by  the  form  :  sin(0±<£)  =  sin  a. 

B.  Determine  the  general  value  of  0  in  the  following  cases : 

1.  sin 0 ±  cos 0  =  V2.  12.  5cos0  -  2 sin0  =  ±  V29. 

2.  sin30±cos30  =  — .  13.  2  cos0  ±  5  sin0  =  ±  V29. 


V2 

3.  sin50±cos50  =  -V2. 

4.  sin 20  ±  cos 26  =  0. 

5.  sin  ra0  ±  cos  nO  =  0. 

6.  sin  mO  ±  cos  nO  =  ±  V2. 


14.  5.32  cos0-2.32  sin  0=2.41. 

15.  2cos0-5sin0  =  2. 

16.  5cos0-2sin0  =  6. 

17.  2cos3  0  +  3cos30  =  3. 

18.  3sin40±4cos40  =  5. 


7.  sin  m$  ±  cos  n$  =  ± __  0   .     .  n      .         .n      . 

y/2  19.  3  sin  4  0  ±  4  cos  4  0  =  4. 

8.  \/3cos0±sin0  =  ±V2.  20.  3sin40  ±  4cos  40=  6. 

9.  V3  sin  0  ±  cos  0  =  ±  V2.  21.  5  sin  5  0  ±  7  cos  5  0  =  3. 

10.  1  +  sin  0  =  V3  cos  0.  22.  5  sin  0  +  2  cos  0  =  5. 

11.  l-cos0  =  sin0.  23.  6  cos0  +  8  sin0  =  9. 


280  PLANE  TRIGONOMETRY.  [§  149 

24.  By  the  aid  of  an  auxiliary  angle,  show  the  changes  in  sign  and 
magnitude  of  sin  0  +  cos  0,  as  0  goes  from  0°  to  360°. 

Ans.  From  1  to  +  V2  to  0  to  -  V2  to  0  to  1,  as  0  passes  from  0°  to 
45°  to  135°  to  225°  to  315°  to  360°. 

25.  Show,  similarly,  the  changes  in  sign  and  magnitude  of : 

(a)  sin  0  —  cos  6.  (c)  sin  6  —  VS  cos  0.  (e)  5  cos  6  —  2  sin  0. 

(b)  sin  6  +  V3  cos  6.     (d)  cos2  $  -  sin2  0.  (/)  3  cos  6  +  4  sin  0. 

26.  Is  there  any  limitation  in  the  possibility  of  solving  by  the  method 
of  auxiliary  angles  ? 

§  149.     Use  of  the  Addition-subtraction  Formulas  in  calculating 
a  Table  of  Sines  and  Cosines. 

sin  (A  +  B)    =  sin  A  cos  B  +  cos  A  sin  B.  (1) 

sin  (J.  —  5)    =  sin  ^L  cos  B  —  cos  J.  sin  B.  (2) 

.-.  sin  (A  +  B)    =2  sin  -4  cos B  -  sin  (^  -  B).  (3) 

.-.  sin  (60°  +  A)  =  sin  ^.  +  sin  (60°  -  4).  (4) 

Formulas  (3),  (4)  may  be  used  in  calculating  a  table  of 
sines ;  (3)  being  used  for  all  angles,  J.,  below  60°,  and  (4) 
to  get  from  these  the  sines  for  all  angles  from  60°  to  90°. 
When  the  sines  are  known,  so  are  the  cosines,  since  cos^l 
=  sin  (90°  -A}. 

For  example,  suppose  the  table  is  to  read  to  every  10",  let 

B  =  10"  and  A  =  10",  20",  30",  40",  etc.,  in  turn.      Then 

(3)  gives : 

v  sin  20"  =  2  sin  10"  cos  10", 

sin  30"  =  2  sin  20"  cos  10"  -  sin  10", 
sin  40"  =  2  sin  30"  cos  10"  -  sin  20", 
sin  50"  =  2  sin  40"  cos  10"  -  sin  30", 
etc.,  etc.,  etc. 

The  table  may  be  started  with  any  angle  whose  sine  and 
cosine  are  known,  when  the  sine  and  cosine  of  the  angle- 
difference  of  the  table  are  known.     For  instance,  let  A  =  30°, 

sin  A  =  0.5000,  cos  A  =  2~  =  0.8660,  for  a  table  reading  to 

minutes.     Let  B  —  V.     Then 

sin  30°  1'  =  sin  30°  cos  1'  +  cos  30°  sin  1'. 


§150] 


THE   SINE   AND   THE   COSINE  IN   UNION. 


281 


Thus  knowing  sin  30°  1',  then  by  (3), 

sin  30°  2' »  2  sin  30°  1'  cos  1'  -  sin  30°, 
sin  30°  3'  =  2  sin  30°  2'  cos  1'  -  sin  30°  1', 
sin  30°  4'  =  2  sin  30°  3'  cos  V  -  sin  30°  2', 
etc.,  etc.,  etc. 

Evidently  from  the  foregoing  it  is  necessary  to  calculate  the 
sine  and  cosine  of  the  angle-difference  of  the  table,  whether 
this  be  10',  1',  10",  or  1",  the  usual  differences. 

§  150.     Calculating  the  Sine  and  Cosine  of  the  Angle-difference 

of  Tables. 

Let  6  be  any  small  angle  in  circular  measure. 

{*!*?  co.p.107 


Then 


We  now  show 


cos  0  <  1. 

sin0>0-— , 
4 

cos  0  >  1  —  — , 
28 


from  which  it  will  follow  that  when  for  the  sine  of  a  small 
angle  its  circular  measure  is  taken,  the  error  is  less  than  one- 
fourth  the  cube  of  the  circular  measure,  and  when  for  the 
cosine  of  a  small  angle  unity  is  taken,  the  error  is  less 
than  one-half  the  square  of  the  circular  measure, 

DAO  (Fig.  145)  is  a  portion  of 
the  unit  circle.  CB,  DB  are  tan- 
gents to  this  circle. 

The  broken  line  DBC> arc  BAG  tf 

.-.  line  BC>  arc  AC.       (1) 

The  number  for  the  arc  AC  is  the 
same  as  that  for  the  angle  AOC. 

.-.  arc  AC  =6. 
Triangles  OEC,  OCB  are  similar. 

CB  =  UC 
"  OC      OE 


282 

PLANE   TRIGONOMETRY. 

But 

00  =  1,  EO  =  sin  0,  OE =  cos  0. 

.-.  OB  =  sin6. 
cos  6 

.'.  by  (1), 

sm0     ^ 

COS0 

,\  sin  0  >  0  •  cos  0. 

[§150 


(2) 

That  is,  the  sine  of  any  small  positive  angle  is  greater  than 
the  product  of  its  circular  measure  and  cosine. 

.    0^0       0  /0n 

•*•  sin2>2C°S2*  (  ^ 

0       0 
But  sin  0=2  sin  -  cos  - . 

A         -i 


sin 


0>2{rosi)oosi 


or, 

sin  0  >  0  cos2  - . 

A 

.:  sin  0>  0^1 -sin2' 

But 

0^   .    0 

2>Sm2' 

02      .  20 

•••4>SmV 

By  (5),  (6), 

sin0>0(l-J), 

or, 

sin0>0-f. 
4 

Also 

cos  0  =  1  —  2  sin2  - 

cos  0  >  1 


-<!> 


(4) 

(5) 

(6) 


or,  cos  0  >  1  —  — . 

A 


§  150]         THE   SINE   AND  THE   COSINE  IN   UNION.  283 

The  circular  measure  of  10' 

=  18raradianS 

=  0.002,908,9,  about. 

.-.  sin  10' =  0.002,908,9, 

with  an  error  less  than  one-fourth  the  cube  of  this  number, 

f0.003)3 
or     <^T-A 

or     <  0.000,000,007,  about, 

or  less  than  1  in  the  eighth  decimal  place. 

Thus,  in  constructing  any  place  table,  up  to  and  including 
a  seven-place  table  of  sines  and  cosines,  we  may  take  sin  10' 
as  equal  to  the  circular  measure  of  10'  for  seven  places  with- 
out appreciable  error. 

Similarly,  sin  1'  =0.000,290,89, 

or,  approximately,  sin  1'  =  0.0003,  or,  3  zeros  and  3. 

Similarly,  sin  1"  =  0. 000,004,848,136,811, 

or,  approximately,  sin  1"  =  0.000,005,  or,  5  zeros  and  5. 

The  error  in  the  above  value  of  sin  1'  is  less  than 
0.000,000,000,007,  and  the  error  in  the  value  of  sin  1"  is 
less  than  0.000,000,000,000,000,03. 

fsin<9<0, 
From  the  formulas    j    •    q  >  q  _  & 

sin  1"  <  0.000,004,848,136,811 ; 
sin  1"  >  0.000,004,848,136,807. 
■%  these  agree  to  13  figures. 

.-.   sin  1"  =  0.000,004,848,136,8, 
which  is  correct  to  13  figures. 

cos  6  =  Vl  -  sin2  d  =  (1  -  sin2  $)*  =  1  -  £  sin2  0, 
approximately  (§  42). 

.-.   cos  1"  =  0.999,999,999,988. 


284  PLANE   TRIGONOMETRY.  [§  150 

02 

Since  cos  6  <  1  and  cos  6  >  1  —  — ,  we  may  set  cos  1"  =  1, 
with  an  error  less  than  |(0.000,004,848)2, 

or    <i(0.000,005)2, 

or     <  0.000,000,000,013. 
Thus  the  first  ten  figures  of  cos  1"  are  9's. 
Similarly,  cos  1'  =  1,  with  an  error  less  than  J(0.0003)2, 

or     <  0.000,000,05, 
and  when  we  set  cos  10'  =  1,  the  error  is  less  than  J  (0.003)2, 
or     <  0.000,005. 
Thus  in  constructing  a  four-place  or  five-place  table  we 
may  assume  cos  10',  or  the  cosine  of  any  angles  less  than 
10',  as  1. 

Similarly,  for  a  four-place,  five-place,  six-place,  seven-place 
table,  we  may  set  cos  1'  =  1,  as  also  for  any  smaller  angle. 
And  for  any  table  of  less  than  eleven  places,  we  may  set 
cosl"  =  l. 

Thus  in  making  a  four-place  table  reading  to  every  10', 
we  may  take 

sin  20' =  2  sin  10'; 
sin  30' =  2  sin  20' -sin  10'; 
sin 40' =  2  sin  30' -sin  20'; 
sin  50'  =  2  sin  40'  -  sin  30'. 
etc.,  etc. 
Thus  also  any  table  for  which  we  may  assume  cos  1"  =  1 
(i.e.  for  any  place   table  below  an  eleven-place  table,  as 
above).  sin  2"  =  2  sin  1"; 

sin  3"  =  2  sin  2"  -  sin  1"  =  3  sin  1"; 
sin  4"  =  2  sin  3"  -  sin  2"  =  4  sin  1"; 
sin  5"  =  2  sin  4"  -  sin  3"  =  5  sin  1". 
In  general,  sin  nn  ==  n  sin  1",  when  n  is  small. 
This  formula  is  but  another  expression  of  the  fact  that 
when  an  angle  is  small  its  sine  may  be  taken  as  equal  to 
its  circular  measure. 


§151]        THE  SINE   AND  THE   COSINE  IN  UNION.  285 

For         sinn"  =  circular  measure  of  nn 

—  n  times  circular  measure  of  1" 
=  n  times  sin  of  1". 
The  error  is  less  than  \  (circ.  meas.)3.     Thus,  for  a  four- 
place  table,  if  this  error  is  not  to  affect  the  table,  it  must 
be  less  than  0.00005. 

Thus,  to  get  the  limit  for  w,  so  that 

sin  n"  =  nsinl" 
for  a  four-place  table, 

{  (n  circ.  meas.  I")3  <  0.00005 ; 
^/Q.0002 
71     0.000005' 
.-.  n  <  about  12,000. 
Therefore  the  formula 

sin  nfr  =  n  sin  1" 
will  hold  up  to  about  n"  =  3°  in  a  four-place  table. 

EXERCISE. 

Observe  this  in  some  four-place  table,  making  some  tests  like 

sin  20'  =  2  sin  10' 

sin  40'  =  2  sin  20' 

sin  60'  =  2  sin  30' 

sin  3°   =  2  sin  1°  30',  etc. 
(Compare  (o),  p.  110). 

§  151.    Useful  Practical  Method  for  Determining  the  Number  of 
Seconds  in  a  Small  Angle  of  Given  Circular  Measure. 

Let  6  =  circular  measure  of  nn, 

.-.  6  =  sin  n"  =  n  sin  1". 

0  0 

"  n     sinl"      0.000,005' 
which,  as  already  pointed  out,  will  hold  for  four  places  up 
to  3°,  and  determine  n  from  0,  or  0  from  n. 

EXERCISE. 

Determine  such  limits  for  five-place,  six-place,  seven-place  tables,  and 
find  the  seconds  in  selected  small  angles  in  circular  measure,  and  vice  versa. 


286  PLANE   TRIGONOMETRY.  [§  152 

§  152.    Basis  for  the  Rule  of  Proportional  Parts  in  a  Table  of 
Sines  and  Cosines. 

sin  (A  +  B)  —  sin  A  cos  B  -f  cos  A  sin  B ; 
cos  ( A  -f  jB)  =  cos  A  cos  B  —  sin  A  sin  B. 

If  /3  =  the  circular  measure  of  J9,  then,  for  as  many  places 
as  we  may  assume 

cos  B  =  1  and  sin  B  =  &  (§  150) 

is  sin  ( A  +  B)  =  sin  A  +  /3  •  cos  J. ; 

and  cos  ( A  +  i?)  =  cos  A  —  ft  •  sin  A 

.-.  sin  (A  +  .#)  -  sin  4  =  £  •  cos  A.  (1) 

cos  (A  +  B)  -  cos  J.  =  -  j3  •  sin  A  (2) 

Formula  (1)  shows  that,  within  the  limitations  noted 
above,  the  difference  in  the  sines  of  two  angles  is  propor- 
tional to  the  difference  in  the  angles. 

Formula  (2)  shows  the  same  thing  for  cosines. 

Let  /3  =  net. 

Then  sin  {A  +  B)  —  sin  A  =  na  cos  A. 

Also  sin  f  A  -\ — J  —  sin  A  =  a  cos  A. 

.*.  sin  {A  +  B)  —  sin  A  =  n  (sin  f  A  -\ —  •)  —  sin  A). 

That  is,  the  sine-difference  for  the  angle -difference  na  is  n 
times  that  for  a,  which  is  the  rule  for  proportional  parts. 

EXERCISES. 

1.  Show  that,  in  any  table  of  sines,  the  number  under  "  D  "  opposite 
any  angle  is  the  cosine  of  that  angle  times  the  circular  measure  of  the 
angle-difference  of  the  table.     Explain  columns  "  P.  P."  of  tables. 

2.  Show  that  in  any  table  of  cosines  the  number  under  "  D  "  opposite 
any  angle  is  the  sine  of  that  angle  times  the  circular  measure  of  the  angle- 
difference  of  the  table. 


§  153]         THE   SINE   AND   THE   COSINE  IN   UNION.  287 

§  153.     Testing  the   Accuracy  of  Calculated  Tables  of  Sines 

and  Cosines. 

The  method  of  §  149  for  calculating  a  table  of  sines  is 
based  upon  additions  and  subtractions  of  results  already 
found.  If  an  error  occurs  at  any  part  of  the  table,  it  affects 
all  the  following  part  of  the  table. 

The  following  tests  of  accuracy  may  be  applied : 


(a)  sinA=l  {VI  +  sin  2  A  -Vl-sin2.A}, 


cos  A  =  J  { Vl  +  sin  2  A  +  VI  -  sin  2  A}, 

A  being  any  angle  less  than  45°.     Test  (a)  is  tedious. 

(6)  sin  (36°  +  A)  —  sin  (36°  -A)  +  sin  (72°  -  A)  -  sin  (72° 
+  A)  =  sin  A. 

For    sin  (36°  +  A)-  sin  (36°  -  A)  =  2  sin  A  cos  36°, 

sin  (72°  -A)-  sin  (72°  +  A)  =  -  2  sin  A  cos  72°. 

.\  sum  =  2  sin  A  (cos  36°  -  cos  72°), 

=  2sin^(V^tl_^=i)  =  sinA 

Test  (6)  is  Euler's  test. 

O)  sin  (54°  +  4)  +  sin  (54°  -  A}  -  sin  (18°  +  A)  -  sin  (18° 
—  A)=  cost!. 

This  is  known  as  Legendre's  test.  It  is  identical  with 
Euler's  test,  from  which  it  is  obtained  by  writing  90°  —  A 
for  A. 

The  student  interested  in  practical  processes  which  have  been  followed 
in  constructing  Logarithmic  Tables  may  consult  Glaisher's  article, 
"Logarithms,"  in  the  Encyclopaedia  Britannica. 

EXERCISE. 

Test  a  few  sines  (cosines)  of  some  table  by  these  formulas. 


288  PLANE  TRIGONOMETRY.  [§154 

§  154.    The  Limiting  Value  of  ^l?  and  of  cos  8  as  0  approaches 

8 

the  Value  Zero. 
In  §  150  it  has  been  shown  that 

sin  6  <  0, 

6Z 
and  sin  6  >  6  —  —• 

4 

sin  6  ^  1       -,  sin  6  ^  -,      02 
...  _<land  —  >l-_. 

Thus  —=—  can  be  made  to  approach  as  near  as  we  please 

u 

to  the  limiting  value  1,  as  6  is  made  smaller  and  smaller. 
This  is  expressed  in  the  form 

5H^  =  l,  when  0  =  0. 
6 
Similarly,  since  cos  6  <  1, 

and  cos  6  >  1  -  ^,  (§  150) 

cos  0  =  1,  when  0  =  0. 

§  155.    Differentiating  the  Sine  and  Cosine. 

If  F(x)  denotes  any  function  of  x,  and  if  x  is  given  the 
value  x  +  Ax,  and  the  difference  F(x  +  Ax}  —  F(x)  is  formed 
and  divided  by  Ax,  then  the  limit  toward  which  the  quotient 

— - — ^ ^— ^  approaches,  when  it  approaches  a  definite 

limit  as  Ax  approaches  zero,  is  called  the  differential  coeffi- 
cient of  F(x)  with  respect  to  x,  also,  the  first  derived  func- 
tion of  F(x)  with  respect  to  x,  and  is  indicated  by  F'(x). 
The  process  here  pointed  out  to  which  F(x)  is  subjected  is 
called  differentiating  F(x)  with  respect  to  x. 

Treating  the  first  derived  function  as  was  the  original 
function  is  called  getting  the  second  differential  coefficient, 
also  getting  the  second  derived  function;  and  so  on.  The 
second  derived  function  is  indicated  by  F"(x),  etc. 


155]        THE   SINE   AND  THE  COSINE  IN  UNION.  289 

For  example, 

(1)  If  F(x)  =  x*, 

F(x  +  As)  «  (x  +  As)2  =  s2  +  2  x  •  As  +  As2. 

^(s  +  As)-.F(s)      0      ,   A 
.-.  — r-^ ^  =  2  s  +  As. 

As 

.-.  F'(x)=2x. 
Similarly,         F"(x)  =  2  and  F"'(x)  =  0. 

(2)  If  ^(s)=sins, 

F(x  +  As)  =  sin  (s  +  As) . 

F(x  +  As)  —  -F(s)  _  sin  (s  +  As)  —  sin  s 

As  As 

0   .   As       (    ,  As\ 
2  sin— -cos(s  +  —  J 


(•+» 


This  can  be  set  in  the  form 

•    As 
ojjj 

^(s  +  As)  -  F(x)  2 

_1 _^ 1_Z  sb  — COS 

As  As 

2 

.    As 
sin  — 

By  the  preceding  section,  — - —  =  1,  when  As  =  0. 

.*.  F'(x)  =  coss.  2 

That  is,  the  differential  coefficient  of  sin  s,  with  respect  to  s, 
is  cos  s,  s  being  in  circular  measure. 
(3)  If  i7(s)  =  coss, 

JP(s  +  As)  =  cos  (s  +  As). 
F(x  +  As)  —  F(x)  _  cos  (s  +  As)  —  cos  s 
As  As 

/        As\  •    As 

-2  sin  As  sin  ^  +  -2")  am—         ,        A 


;in  f 


sin    s  + 


As  As 

2 
.-.  .F'(s)  =  -sin(s). 

That  is,  the  differential  coefficient  of  cos  x  with  respect  to  x 
is  —  sin  s,  s  being  in  circular  measure. 


290 


PLANE   TRIGONOMETRY. 


[§155 


(4)  We  thus  have : 

When  F(x)  =  sin  #, 

F'(x)  =  cos  a?, 

F"(x)  =  —  sin  as, 

Ff"(x)=-cosx, 

Fw(x)  —  sin  x, 

and  then   a   repeat  without 
end. 


When  F(x)  =  cos  x, 

F,(x)=-smx, 

F"(x)  =  —  co8z, 

F"f(x)=smx, 

Fiv(x)=cosx, 

and   then   a  repeat  without 
end. 


(5)  If,  now,  we  denote  by  ^T(0),1T,(0),  .F"(0),  etc.,  the 
special  values  which  these  functions  assume  when  x  =  0, 
we  have 


When  F(x)  =  sin  x, 
^(0)=0, 
J"(0)=1, 

^(0)=0, 

2^(0)=  0, 

and  a  repeat  without  end. 

(6)  If  F(x)  =  xn, 

where  n  is  a  positive  integer, 
JP(>  +  Ax)  =  O  +  Az)w 

=  xn  +  w^"1  Az  +  -  '  w 0~  1  xn~2  Ax~2  4-  etc., 
assuming  the  binomial  theorem  for  positive  integral  indices. 


When  F  (x)  =  cos  x, 

^(0)=i; 

JF'(0)  =  0, 
and  a  repeat  without  end. 


jTg  +  Agj  -  JPCa;)  _ 


Ax 


=  nxn~x  -f  powers  of  Ax. 


and  if 


.%  F'(x)=nxn~l. 
.-.  if  ^0)=^,  _F'0)=3a;2, 
F(x)  =  aa,-4,  jF'O)  =  4  a^,  etc. 


§  156]         THE   SINE   AND  THE   COSINE  IN  UNION.  291 

§  156.   The  Sine-series  and  Cosine-series. 
Assuming  that  sin  0  can  be  expressed  in  powers  of  0,  let 
_F(0)  =  sin  0  =  a  +  bd  +  eft  +  d0*  +  efr  +/05  +  etc. 
.-.  F'(ff)  =  cos  0  =  b  +  2  c0  +  3  d&  +  4  <?03  +  5/0*  +  etc., 
jF"(0)  =  -  sin  0  =  2  c  +  2  •  3  •  <?0  +  4  •  3  efl2  +  5  •  4/03  +  etc., 
JF"? (0)  =  -  cos  0  =  2  •  3  •  d  +  4 ' •  3  •  2  e0  +  5  •  4 .  3/02  +  etc., 
j^(0)=  sin  0  =  4-3.2.e  +  5-4.3-2/0-h  etc., 
F\d)  =  cos  0  =  5  •  4  •  3  .  2/+  6.5.4.3.  2^0  +  etc. 

etc. 

Assuming  that  these  relations  hold  for  all  values  of  0, 
let 


0  =  0. 

. 

•.  a- 

=  sin(0)=0, 

V 

b: 

=  cos(0)=l, 

2C: 

=  -  sin  (0)  = 

0.     .-.  <?= 

:0. 

2 . 

3d: 

=  -cos(0)  = 

-1.     .-.  d  = 

1 

ft* 

:0. 

4 

•3 

•  2e-. 

=  sin(0)  =  0. 

.-.  e  = 

5.4 

•3 

•2/= 

=  cos(0)=l. 
etc. 

•••/= 

1 

'\1 

.\  Setting 

those  values  of  a,  b,  c, 

etc.,  in  the  selected  series, 

sin 

0  = 

=  0- 

03        05 

"ft     ft      '"' 

to  infinity. 

Similarly, 

cos 

0  = 

=  1- 

02        04 

"ft  H  "' 

to  infinity. 

These  are  the  series  heretofore  given  as  expressing  what 
is  really  meant  by  the  sine  and  cosine  as  related  numerically 
to  a  given  angle  0,  0  being  in  circular  measure. 


292  PLANE   TKIGONOMETRY.  [§  157 

§  157.     The  Sine-series  and  Cosine-series  are  Convergent. 

A  series  is  said  to  be  convergent  when  the  larger  and 
larger  the  number  of  terms  summed,  the  nearer  and  nearer 
some  definite  finite  quantity  is  approached. 

The  sum  of  the  series, 

1_1  +  1_1  +  1_1  +  1,  etc.  ... 

is  not  infinite.  It  is  either  +  1,  or  0,  according  to  the 
number  of  terms  taken.  Such  a  series  is  not  convergent 
under  the  definition.  Many  algebras  give  a  wrong  impression 
as  to  convergency  of  series,  restricting  divergent  series  to 
those  whose  sum  is  infinite. 

If  the  terms  of  a  series  are  real,  and  if  from  some  term 
within  a  finite  number  of  terms  from  the  beginning-term 
they  diminish  numerically,  with  alternating  signs,  the  series 
is  convergent,  if  the  order  of  its  terms  is  not  disturbed. 

Let  the  rth  term  be  a  positive  term  beyond  which  the 
terms  alternate  in  sign  and  diminish.  Call  this  term  An  and 
the  sum  of  the  preceding  terms,  Sr.     The  series  is 

S  =  Sr  +  (Ar  -  Ar+l)  +  (Ar+2  -  Ar+3)  +  (      )  - 
or   S  =  Sr  +  Ar  -  (Ar+1  -  Ar+2)  -  (Ar+3  -  Ar+i)  -  (      ) .... 

Since  the  bracketed  terms  are  all  plus  in  the  first  arrange- 
ment and  all  minus  in  the  second,  S  is  more  than  Sr  and  less 
than  Sr  +  Ar.  Thus,  the  true  sum  of  the  series  differs  from 
Sr  by  less  than  Ar.  Evidently,  then,  by  increasing  r,  the 
nearer  is  a  definite  sum  reached.  The  quantities  Sr,  Sr+1, 
Sr+2,  etc.,  have  thus  the  limit  S,  which  is  the  sum  of  the 
series. 

That  the  order  of  the  terms  of  such  a  series  must  not  be 
disturbed  when  the  convergency  depends  on  the  signs,  was 
first  pointed  out  by  Riemann.  For  we  may  make  such  a 
series  approach  any  sum  whatever  if  we  change  the  order  of 
terms  appropriately.  We  may  add  up  positive  terms  until 
a  sum  larger  than  the  selected  sum  is  reached.     Then  take 


§  158]         THE   SINE  AND   THE   COSINE   IN   UNION.  293 

negative  terras  until  a  sum  less  than  the  selected  sum  is 
reached.  Then  again  add  positive  terms  until  the  selected 
sum  is  passed.  Then  attach  negative  terms,  to  bring  the  sum 
again  below  the  selected  sum,  and  so  on.  Evidently,  the 
more  and  more  terms  we  take  in  this  manner,  the  nearer  we 
approach  the  selected  sum. 

In  the  sine-series  the  ratio  of  the  (r  +  l)th  term  to  the 
rth  is  —  62/2  r  (2  r  +  1),  which  is  numerically  less  than  1 
so  soon  as  r  >  0,  or  r  >  JV#2  +  \  —  \.  Thus  the  sine-series 
is  convergent.  So,  likewise,  the  cosine-series.  Both  of 
these  series  remain  convergent  if  all  their  terms  are  taken 
positive,  for  each  would  then  be  a  part  of  the  series  for  e0 
(§  8),  a  convergent  series.  A  part  of  a  convergent  series, 
all  of  whose  terms  are  positive,  must  be  convergent.  Thus 
the  convergency  of  the  sine-series  and  cosine-series  is  not 
dependent  on  the  signs  of  the  terms,  and  is  consequently 
independent  of  the  order  in  which  the  terms  are  taken. 
These  series  are  what  is  termed  absolutely  convergent.  The 
case  is  very  different  with  the  series  for  loge  (1  +  z),  §  9, 
where  the  convergency  is  dependent  on  the  signs,  and  that 
series  converges  to  the  logarithm  only  when  the  order  of 
terms  is  held  as  given  in  §  9.  For  information  more  in 
detail  on  series,  consult  Osgood's  "Infinite  Series." 

§  158.     Sine  and  Cosine  Relations  for  the  Angles  of  a 
Triangle. 

A  +  B  +  (7=180°.  (1) 

.:A  +  B  =  180°-  0.         .'.A  +  B=90°-Q. 

2  2 

.\  sin  (JL  +  B)  =  sin  0;  cos  (A  +  B)  =  —  cos  0; 

.    A  +  B  n  A  +  B       .    n 

sin  — g —  =  cos  (J  ;  cos  — j- —  =  sin  C ; 

with  similar  relations  for  any  pair  of  angles. 

From  these  relations,  together  with  the  addition-multipli- 
cation, etc.,  relations,  can  be  deduced  many  others. 


294  PLANE   TRIGONOMETRY.  [§  158 

(1)  To  prove 

sin  2  A  +  sin  2  B  -f  sin  2  (7  =  4  sin  J.  sin  2?  sin  0. 


sin2J.  +  sin2^  +  sin  2  (7=  sin  2^4  +  sin  aB  +  sin  2  (7 
=  2  sin  (A  +  .£)  cos  (.4  -  B)  +  2  sin  0  cos  (7 
=  2  sin  0  cos  (J.  -  B)  +  2  sin  (7  \  -  cos  ( J.  +  B)  I 
=  2  sin  C{cos  (A  -  B)  -  cos  (A  +  B)\ 
=  2  sin  (7(2  sin  ^4  sin  B) 
=  4  sin  J.  sin  B  sin  C. 

(2)  To  prove 

cos  A  +  cos  i?  —  cos  C  =  —  1  +  4  cos  —  cos  —  sin  — . 

AAA 


cos  J.  +  cos  B  —  cos  (7  =  cos  A  +  cos  B  —  cos  (7 
=  2  cos2|-  1  +  2  sin  <Lz*  sin  £±J? 

=  2cosf  sin(^±*)_  1  +  2si„^cos| 
At  .    0+  B  .     .    0-By 


o       A?0   .    C       B\      ., 

=  2  cos—   2  sin— cos  —    —  1 

2  V  2        2/ 

ABC 

=  —  1  +  4  cos—  cos—  sin—. 

AAA 


EXERCISES. 

(^  +  5  +  C  =  180°.) 

1.  am*  A  +  sin2J5  +  sin2  C  =  2  (1  +  cos  A  cos  5  cos  C). 

2.  sin  2  A  +  sin  2  5  -  sin  2  C  =  4  cos  4  cos  5  sin  C. 

3.  cos  2  A  +  cos  2  £  +  cos  2  C  =  -  1  -  4  cos  A  cos  5  cos  C. 

4.  cos2^4  +  cos2£-cos2  C  =  1  -  4  sin  J  sin  B  cos  C. 

5.  sin  A  +  sin  £  -f  sin  C  =  4  cos  —  cos—  cos—. 

2         2         2 


§  159]         THE   SINE  AND   THE   COSINE  IN   UNION.  295 

ABC 

6.  sin  J.  +  sin  5  —  sin  C  =  4  sin—  sin—  cos— . 

2        2        2 

ABC 

7.  cos  A  +  cosjB  +  cos  C  =  1  +  4  sin—  sin  — sin— . 

2        2        2 

8.  sin2  A  +  sin2  B  -  sin2  C  =  2  sin  A  sin  B  cos  C. 

9.  cos2  4  +  cos2  B  +  cos2  C  =  1  -  2  cos  A  cos  5  cos  C. 

10.  cos2  4  +  cos2  B  -  cos2  C  =  1  -  2  sin  A  sin  £  cos  C. 

11.  sin2  ^  +  sin2!  +  sin2^  =  1  -  2  sin-  sin-  sin^. 

Z  2.  1  2         2         2 

12.  sin24+  sin2!-  sin2^  =  1  -  2  cos-cos^  sin^. 

2  2  2  2        2        2 

13.  sin  (.4  +  2£)+sin(fl  +  2C)  +  sin(C  +  2^) 

=  4sini-^sin£^sin^i. 
2  2  2 

ARC 

14.  sin^  +  sin^+  sin- 

=  1  +  4sin180O--4Sin180O--Bsin180°-C. 
4  4  4 

15.  sin  {A  +B-  C)  +  sin(J5+  C-A)  +  sin(C  +  A  -  B) 

=  4  sin  A  sin  B  sin  C. 


§  159.  Relations  connecting  the  Sides  of  a  Triangle  with  the 
Sines  and  Cosines  of  its  Angles. 

We  have  shown  that  in  any  triangle, 

a  b  c  ^.x 

sin  A      sin  B     sin  0  ' 

a  =  b  cos  0  -f-  c  cos  B,    (ii)  with  similar  equations  for  5,  c ; 
a2  =  b2  +  c2—  2  be  cos  J.,  (iii)  with  similar  equations  for  5,  <? ; 


sin  —  =  \/i =p ^i,  (iv)  with  similar  equations  for  ■— ,  —  . 

A        *  be  2    2' 

cos  —  =  \       r  a  •>  Cv)  w^n  similar  equations  for  —,  —  ; 

A        A 

sin  A  =  2  sin  —  cos  — ,       (vi)  with  similar  equations  for  B,  C. 
A  A 


296  PLANE   TRIGONOMETRY.  [§  159 

From  these  relations,  and  the  treatment  of  the  angles  as 
in  the  preceding  section,  follow  many  other  relations.  For 
example,  from  (i),  a  _  sin  A 

b      sin  B 

0  .   A  +  B       A-B 

.     -^      2  sin  — ! — ■  cos  — - — 

fl  +  o_sini  +  sin5_  2  2 

a  —  b  ~~  sin  A  —  sin  B  ~  0   .    A  —  B        A  +  B' 

2  sin  — - —  cos  — j- — 

2  2 

,     .  n   .    A-B        A  +  B     ,       ,.    .    A+B        A-B 

.-.  (a  +  o)  sin — - —  cos  — - —  =  (a  —  b)  sin  — ^ —  cos  — - — . 


EXERCISES. 


1.    (b  +  c)sm-=acosB 


2. 


2  2 

A         B 

2  cos  —  cos  - 
a  +  b  +  c  2         2 


.    C 

sin  — 

2 


3.  (b  -  c)  cos  -  =  a  sin  ^  ~  C. 
K         J        2  2 

4.  a(cos  B  +  cos  C)  =  2(6  +  c)  sin2-. 

5.  a(cos  £  -  cos  C)  =  2(c  -  &)  cos2—. 

6.  a2  +  62  +  c2  =  2(ab  cos  C  +  be  cos  .4  +  ca  cos  £). 

7.  c2  =  (a  -  6)2  cos2-  +  (a  +  &)2sin2-. 

8.  a  sin  (5  -  C)  +  b  sin  (C  -  A)  +  c  sin  (.4  -  B)  =  0. 

9.  — ^— .sin(5-  C)=   -  b      .Sm(C-A)  =  —£ sin  (A-B). 

b2-c2         K  J     c2-a2         K  J      a2-b2         K  ' 

10.  a  Bm-  sin  ^-=-^  +  &sin-  sin  C  ~A  +csin-sin  A^J*  =  o. 

2  2  2  2  2  2 

11.  a2(cos2J5  -  cos2C)  +  &2(cos2C  -  cosM)  +  c2(cos2^  -  cos2£)  =  0. 

12.  £Lzig?  sin  2  A  +  gi^-g-2  sin  2  £  +  -  ~  ^  sin  2  C  =  0. 

a2  62  c2 

13.  a  +  6  +  c  =  (a  +  b)  cos  C  +  (b  +  c)  cos  ^4  +  (c  +  a)  cos  B. 

14.  sin  (5  -  C)  =  ^-=-£-2  sin  ^. 


CHAPTER  VIII. 


3p 

^K 

N 

\ 

^ 

^ 

V                                p  «•"' 

v                                  _^^"^ 

^-                  ^^^ 

M                  \<r*^           ' 

1/                -<  -  -N                   ilf 

J[          „2_         *5                 M 

±                              sw 

*£                                       -*« 

Z?t                                                 5* 

i                                                             -iH- 

j*                                                             * 

Fiq.  146. 


THE   TANGENT   AND   RECIPROCAL   TANGENT    (COTAN- 
GENT) OF  ANGLES. 

§  160.   The  Tangent  of  an  Angle  is  the  ratio  of  the  ordinate 
of  any  point  on  its  terminal  to  the 
abscissa  of  the  same  point. 

In  Fig.   146   the   tangent   of 
any  angle  whose  terminal  is  OP 
.    MP 
1S  OM' 

LABORATORY   EXERCISES. 

1.  Lay  out  with  the  protractor  a 
given  angle.  Measure  five  pairs  of 
ordinates  and  abscissas  for  this  angle, 
one  abscissa  being  a  unit  (inch  or 
foot).  Divide  to  as  many  figures  as  the  plan  of  measurement  will  justify 
and  compare  results  with  each  other  and  with  the  table-tangent  for  the 
same  angle. 

2.  Lay  out  five  different  angles.  Measure  for  each  an  ordinate  and 
abscissa.     Divide.     Compare  with  the  table-tangents. 

3.  Lay  out  the  following  points  on  coordinate  paper.  Calculate  the 
tangents  to  one  or  two  decimal  places.  Measure  the  angles  with  the 
protractor.  Compare  with  the  table-tangents  for  the  same  angles : 
(2,3);  (-2,-3);  (2,-3);  (-2,3);  (4,3);  (.5,-4);  (-6,7); 
(3,  -4);  (12,  -5.2);  (-15,  -8.3);  (24,7.3);  (-21,20);  (35,  -12). 

§  161.     The  Sign  of  the  Tangent. 

When  the  ordinate  and  abscissa  have  the  same  sign,  the 
tangent  is  positive ;  otherwise  it  is  negative.  Thus  the  tan- 
gents of  all  angles  whose  terminals  are  in  the  first  or  third 
quadrant  are  positive ;  those  for  all  angles  of  the  second  or 
fourth  quadrant  are  negative. 

297 


298 


PLANE   TRIGONOMETRY. 


[§162 


EXERCISE. 

Determine  the  signs  of  the  tangents  of  the  angles  in  Exs.  1,  2  under 
the  sine,  §  61. 


§  162.    Angles  with  the  Same  Tangent  (Periodicity). 

(1)   All  angles  with  the  same  terminal  have  the  same 
tangent.  .    tan  (2  n  .  18Qo  +  j^  m  tan  ^ 

tan  (2  n  •  ir     +  0)   =  tan  6. 


P 

JJ 

t 

y 

/ 

y 

— 

M; 

y_£ 

0  , 

-M 

r_ 

- 

t* 

M 

**    - 

?* 

/ 

s 

p, 

Fig.  147. 


Fig.  148. 


(2)  All  angles  with  opposite  terminals  have  the  same 
tangent. 

For  then  when  the  moduli  are  equal,  oppositely  equal  are 
the  abscissas,  and  so  also  the  ordinates.     If  0P1  =  —  &P2, 


M1P1  =  M2P2 

-  0Ml      0M2  ' 


If  any  terminal  is  located  by  the  angle  A,  it  is  also  located 
by  the  angle  2  n  •  180°  +  A,  while  the  opposite  terminal  is 
located  by  (2  n  -f  1)  180°  +  A.  Thus  either  an  even  or  an 
odd  number  of  multiples  of  180°  may  be  added,  positively  or 
negatively,  to  an  angle  without  altering  its  tangent ; 

or,  tan  (n  ■  180°  +  A°)  =  tan  A°,  tan  (nir  +  0)  =  tan  0, 
where  n  is  any  positive  or  negative  integer. 

Thus  the  tangent  is  a  periodic  function  with  the  period  it. 
What  is  the  period  for  the  sine,  -cosine,  secant,  cosecant  ? 


163] 


THE   TANGENT   FAMILY. 


299 


EXERCISES. 

1.  Determine  ten  angles  on  diagrams  which  have  the  same  tangent  as 
30°,  five  of  them  having  the  same  terminal  and  five  the  opposite  terminal. 

2.  Do  the  same  for  -  60°.     Solve  sin  6  6  =  tan  3  0. 

3.  Do  the  same  for  -  and  —  -  •     Solve  cos  3  0  =  tan  3  0. 

4  4 

4.  Find  the  angles  which  satisfy  the  relation,  tan  9  $  —  tan  8  6. 

§  163.     Angles  with  Opposite  Tangents. 

(1)   Terminals  symmetric  to  the  horizontal  give,  for  equal 
moduli,  oppositely  equal  ordinates,  while  terminals  symmet- 


x^ 

^i                            lt-7 

T\                                 A 

i             ss                   7 

5^                                                          7 

^^                                               5                      7. 

^^                                              S            7 

^   ^                                            s     ? 

-9.^  -         --           -  -    -    _sz__ 

^t-                   .]/.                     M.                      C                   M, 

^^^^ 

"^^ 

^> 

2TS 

Fig.  149. 


Fig. 150. 


ric  to  the  vertical  give,  for  equal  moduli,  oppositely  equal 
abscissas.  In  the  first  case  the  abscissa  is  unchanged,  and 
in  the  second  case  the  ordinate  is  unchanged. 


Or, 


and 


MPt  -MPX_  _MPA 

OM  OM            OM' 

M,P,=  M,P,  m     M^ 

OM1  "  -OMx          OM2 


Therefore, 

(a)  The  tangent  of  a  negative  angle  is  the  negative  tan- 
gent of  the  corresponding  positive  angle. 

(b)  The  tangent  of  the  supplement  of  an  angle  is  the 
negative  of  the  tangent  of  the  angle. 


300  PLANE   TRIGONOMETRY.  [§  163 

Or,  tan^°  =  -tan(-^°);  (1) 

tan  0    =  -tan(-0);  (2) 

tan  A°  =  -  tan  (180°  -  A9) ;  (3) 

tan0    =- tan(7r-6>);  (4) 

tan  A°  =  -  tan  (n  .  180°  -  A°) ;  (5) 

tan  0     =  —  tan  (n  •  7r  —  0) ;  (6) 

tan  O  •  180°  +  4°)  =  -  tan  (m  •  180°  -  A°) ; 

tan  (w  •  7r  +  0)  =  —  tan  (m  "tt  —  0). 

EXERCISES. 

1.  Diagram  five  angles  whose  terminals  are  symmetric  to  the  hori- 
zontal with  the  terminal  for  30°  and  having  tangent  oppositely  equal  to 
that  of  30°.     Solve  sin  5  0  =  -  tan  10  0. 

2.  Diagram  five  angles  having  their  terminals  symmetric  to  the  ver- 
tical with  the  terminal  of  135°  and  with  their  tangent  oppositely  equal 
to  that  of  135°.     Solve  tan  6  0  =  -  tan  8  6. 

§  164.     Tangents  of  all  Angles  are  First  Quadrant  Tangents. 

It  is  left  for  the  student  to  draw  conclusions  similar  to 
those  made  for  the  sine  in  §  64,  and  report  them.  Give  also 
rules  for  the  terminal  positions,  similar  to  those  in  the  same 
article. 

EXERCISES. 

Take  those  in  §  64,  replacing  the  word  "  sine  "  by  "  tangent." 

§  165.     Construction  of  the  Terminals  for  a  Given  Value  of  the 

Tangent. 

Example.     Draw  the  terminals  when  tan  A  is  f . 
Construct  on  coordinate  paper  the  point  (3,  2)  and  join  it 
with  the  origin. 

Draw  also  the  terminal  opposite  to  this  line. 

EXERCISES. 

Construct  the  terminals  corresponding  to  the  following  values  of  the 
tangent :  f ,  -  §,  } ,  -  §,  .3,  3.7,  33.7,  0,  oo.  Construct  terminals  for  table 
tangents.    Measure  the  angles  and  compare  with  table. 


§  167]  THE   TANGENT  FAMILY.  301 

§  166.     Some  Angles  for  which  the  Tangent  is  readily  calculated 
without  the  Use  of  Tables. 

These  are  the  same  as  those  for  which  the  sines  were  calcu- 
lated in  §  75.  It  is  left  as  an  exercise  for  the  student  to 
establish  the  following  results,  using  diagrams  : 

tan    45°  =  1;  tan  135°  =  -  1 ; 

tan    30°  =  ^;         tan  330°  =  -^; 
o  3 

tan    60°  =  V3  ;         tan  300°  =  -  V3  ; 
tan  120°  =  -  V3  ;      tan  240°  =  V3  ; 

tan  150°  =  -  ^P ;     tan  210°  =  ^ ; 
o  3 

tan  225°  =  1 ;  tan  315°  =  -  1. 

EXERCISES. 

1.  Give  the  general  solution  of  the  equations,  tan2  x  =  1 ;  tan2  x  =  \ ; 
tan2  x  =  3;  tan2  x  =  0 ;  tan2 a:  =  sin2 a;;  tan2 a:  =  3  cos2 a: ;  3  tan2 a:  =  sec2 a:. 

2.  tan2a;  -  H±^l  tana:  +  ^1=0.  4.   tana: L_  =  1  _  ^1. 

3  3  tana;  3 

3.  tan2 a:  -3  ~  ^3  tan  x  --^  =  0.  5.   tan2a:  +  8  tana:  +  7  =  0. 

3  3 


§  167.   Value  of  the  Tangent  when  the  Terminal  is  on  a 
Border  Line  of  the  Quadrants. 

If  the  point  P  of  the  terminal  OP  is  supposed  to  describe 
a  circle  of  radius  r  about  0,  then  as  OP  approaches  the  posi- 
tion of  the  initial  line  to  the  right,  OM  approaches  the  value  r, 

MP 

while  MP  approaches  the  value  zero.     The  value  of  ——- 

OM 

thus  approaches  the  value  zero.  This  is  expressed  by  say- 
ing tan  0°  =  0.  Similarly,  when  OP  approaches  the  initial 
line  to  the  left  of  the  origin,  OM  approaches  the  value  —  r 
and  MP  approaches   the   value   zero.     The   tangent   thus 


302  PLANE  TRIGONOMETRY.  [§167 

approaches  the  value  zero  when  the  angle  approaches  180°. 
Thus  tan  180°  =  0. 

When  OP  approaches  the  upright  vertical  from  the  right, 
MP  approaches  the  value  r  and  OM  approaches  zero  from 

MP 

the  positive  side.  Thus  — — -  becomes  infinitely  great,  posi- 
tively. If,  however,  OP  approaches  the  upright  position  from 
the  left,  MP  approaches  the  value  r  while  OM  approaches 
the  value  zero  from  the  negative  side.  Thus  the  tangent 
becomes  infinitely  great,  negatively.  It  is  thus  impossible  to 
determine  the  value  of  the  tangent  for  upright  positions  of 
the  terminal.     It  is  customary  to  write 

tan  90°  =  +  oo. 
And,  similarly,  tan  270°  =  —  oo. 

This  is  a  short  way  of  saying  that  if  the  terminal  is 
upright,  almost,  with  a  slight  tilt  to  the  right,  the  tangent  is 
a  very  large  positive  number,  while  if  the  terminal  is  down- 
right, almost,  with  a  slight  passage  beyond  the  position 
denoted  by  270°,  the  tangent  is  a  very  large  negative  num- 
ber. Thus,  if  we  look  on  the  terminal  as  moving,  counter- 
clockwise, about  the  origin,  there  is,  when  *it  passes  the 
positions  denoted  by  90°  and  270°,  a  sudden  spring  in  the 
value  of  the  tangent  from  plus  infinity  to  minus  infinity. 
Thus  in  the  function  tan  x  there  is  what  is  called  a  break  in 
continuity  of  value  when  x  is  90°  or  270°,  or,  in  general, 

when  x  is  (2  n  -+- 1)90°,  or  (2  n  +  1)—,  n  being  any  positive 
or  negative  integer. 


§  168.     Line  Picture  of  the  Tangent. 

MP 
If  in  tan  0  =  -pr=-f->  OM  is  taken  as  unity,  tan  6  is  MP. 

Thus,  on  any  scale  on  which  OM  is  taken  to  represent  1, 
MP  will  represent  the  value  of  the  tangent  of  all  angles 
corresponding  to  the  position  OP  of  the  terminal. 


168] 


THE   TANGENT   FAMILY. 


303 


If  we  imagine  the  point  P'  to  describe  the  unit-circle,  and 
if  through  each  position  of  P'  on  the  unit-circle  we  imagine 
a  line  OP'  prolonged  to  meet  the  tangent  at  M  in  the  point  P, 
we  shall  have  for  each  such  position  of  P'  a  line  picture  of 

Pi 


XT       X 

±        2L 

^      '  ^^M^ 

s           -  ^ 

-/                       z  V- 

J                  -**-    V 

^                       -2   -     \ 

1                       -7             \ 

-                   -7             -it 

/                 u 

c 

3                                  f 

L                                                             A 

V                                 2 

s                         / 

S^                               A 

**.-              ~Z2Z 

y 


i\ 


M 


Fig.  151. 


Fig.  152. 


the  corresponding  tangent  denoted  by  MP.  Thus,  tangent 
of  angle  M0P{  is  MPX ;  tangent  of  angle  MOP2'  is  MPV 
and  so  on. 

When  OP-  is  coincident  with  OM,  the  angle  is  zero,  and 
so  is  MP.  As  the  angle  increases  from  zero  to  90°,  the  point 
Pf  describing  the  circle  counter-clockwise,  the  point  P  ascends 
the  line  MP,  and  the  tangent  increases.  When  the  terminal 
is  almost  upright,  the  point  Jr  is  high  up  on  the  line,  cor- 
responding  to'  tan  90°  = +  00, 

in  the  preceding  section.  When  OP1  passes  90°,  P  passes  from 
high  up  on  the  line  to  loiv  down  on  the  line,  corresponding  to 
the  spring  from  plus  infinity  to  minus  infinity  noted  in  the 
preceding  section,  when  the  angle  passes  the  value  90°. 
When  OP'  moves  from  90°  to  180°,  the  motion  of  the  cor- 
responding point,  P,  on  the  line  is  from  negative  infinity  on 
the  line,  up  to  the  point  M.  As  OP',  goes  from  180°  to 
270°,  P  goes  on  up  the  line  from  M  to  plus  infinity.  There 
is  again  the  spring  from  plus  infinity  to  minus  infinity  as 


304  PLANE   TRIGONOMETRY.  [§168 

OP'  passes  through  the  position  270°.  That  is,  as  OP'  passes 
through  270°  the  corresponding  point  P  leaps  from  high  up 
on  the  line  to  low  down  on  the  line.  As  OP'  goes  from  270° 
back  to  the  starting  point,  P  moves  from  minus  infinity 
to  zero  on  the  line,  that  is,,  from  low  down  on  the  line  up 
to  the  point  M. 

The  line-diagram  teaches  clearly  the  following  propositions  : 

(a)    tan  6  =  —  tan(—  0). 

(5)  tan  (180°  +  A°)  =  tan  A*. 
O)    tan  (180°  -A°)=-  tan  4°. 

(d)  tan  0°  =  0  =  tan  360°. 

(e)  tan  180°  =  0  =  tan  (  -  180°). 
(/)  tan  ft .  180°  =  0. 

(#)  tan  90°  =  -f  oo  (as  already  agreed  on). 

(K)    tan  270°  =  —  oo  (as  already  agreed  on). 

(i)    tan  (4^-f- 1)90°  =  oo  (as  already  agreed  on). 

(/)    tan  (An  +  3)90°  =  —  oo  (as  already  agreed  on). 

(6)  As  the  angle  passes  from  0°  to  90°,  or  from  180°  to 
270°,  the  tangent  is  positive  and  an  increasing  function  of 
the  angle.  As  the  angle  passes  from  90°  to  180°,  or  from 
270°  to  360°,  the  tangent  is  negative,  but  an  increasing  func- 
tion of  the  angle.  The  tangent  is  always  an  increasing  func- 
tion, except  on  the  instantaneous  leap  from  +  oo  to  —  oo. 

(?)  The  tangent,  like  the  sine  and  cosine,  is  a  periodic 
function  of  the  angle,  the  period  being  360°,  or,  in  circu- 
lar measure,  2  7r.  It  has  also  the  period  tt,  since  tan  6  = 
tan  (6  -f  mr),  for  all  integral  values  of  n. 

(m)  For  all  angles  whose  terminals  are  in  the  first  or  third 
quadrant,  the  tangent  is  positive ;  for  all  angles  whose  ter- 
minals are  in  the  second  or  fourth  quadrant,  the  tangent  is 

negative. 

LABORATORY  EXERCISE. 

Draw  a  diagram  like  the  preceding,  with  the  radius  as  one  foot. 
Measure  ten  angles  and  tangents.  Compare  results  with  the  table- 
tangents  for  the  same  angles.     Make  a  graph  for  the  tangent. 


§169]  THE  TANGENT  FAMILY.  305 

§  169.     Some  Simple  Calculations  with  the  Tangent. 

Since  the  tangent  is  the  ratio  of  the  ordinate  to  the 
abscissa,  when  any  two  of  the  three  quantities,  ordinate, 
abscissa,  tangent,  are  given,  the  third  can  be  calculated. 

(a)  Given  ordinate  and  abscissa. 

Tangent  equals  ordinate  divided  by  abscissa. 
(6)  Given  abscissa  and  tangent. 

Ordinate  equals  abscissa  multiplied  by  tangent. 
(<?)    Given  ordinate  and  tangent. 

Abscissa  equals  ordinate  divided  by  tangent. 

These  relations  follow  at  once  from  the  definition  of  the 
tangent :  , .     , 

tangent  =  2^125*2. 
abscissa 

Make  diagrams  and  test  by  measurement  the  following 

EXERCISES. 

(Tables  not  to  be  used  wbere  30°,  60°,  45°  occur.) 

1.  When  the  ordinate  is  23  and  the  abscissa  is  37,  calculate  the  tan- 
gent to  two  decimal  places. 

2.  When  the  tangent  is  1.32  and  the  abscissa  is  24.5,  what  is  the  ordi- 
nate to  three  significant  figures? 

3.  When  the  tangent  is  23.41  and  the  ordinate  is  34.67,  what  is  the 
abscissa  to  four  significant  figures  ? 

(Use  the  shortened  process  of  division  and  multiplication  in  Exs.  1, 
2,  and  3.) 

4.  A  vertical  stick  a  feet  long  casts  a  shadow  a  feet  long,  what  is  the 

sun's  angular  elevation?    What  for  shadow  lengths,  aV3,  ^— —  ? 

3 

5.  If-  a  vertical  flagstaff  subtends  an  angle  of  30°  at  a  horizontal  dis- 
tance of  150  feet  from  its  foot,  show  that  its  height  is  50  V3.  Calculate 
this  to  three  significant  figures. 

6.  Calculate  in  radicals  and  also  to  three  significant  figures,  if  the 
angle  in  Ex.  5  is  60° ;  45°. 


306  PLANE  TRIGONOMETRY.  [§  169 

7.  Show  that  tangents  can  be  used  to  solve  the  following  problem ; 
A  man  wishing  to  determine  the  height  of  a  church  spire,  found  its 
angles  of  elevation  from  two  points,  distant  apart  100  ft.,  in  the  same 
horizontal  line  with  the  base  of  the  spire,  to  be  45°  and  60°,  from  which 
he  found  the  height  of  the  spire  to  be  50(3  +  V3).  Calculate  this  to 
three  significant  figures.  How  far  are  the  two  points  from  the  base  of 
the  tower  ? 

8.  From  the  top  of  a  cliff,  200  ft.  high,  the  angles  of  depression  of 
the  top  and  bottom  of  a  tower  are  30°  and  60° ;  use  tangents  to  calculate 
the  height  of  the  tower. 

9.  The  angle  of  elevation  of  a  tower  due  north  from  a  certain  point 
is  30°.  At  a  distance  300  ft.  due  west  from  the  first  point  the  elevation 
is  60°.  What  is  the  height  of  the  tower  in  radicals  and  to  three  signifi- 
cant figures? 

10.  What  is  the  angular  elevation  of  the  sun  when  the  shadow  of  a 
vertical  stick  is  V5  times  its  height  ? 

11.  At  a  certain  point  the  tangent  of  the  angle  of  elevation  of  a  tower 
is  f ,  and  at  a  point  32  ft.  nearer  the  tower  in  a  line  directly  toward  the 
tower  from  the  first  point,  the  tangent  of  the  angle  of  elevation  of  the 
tower  is  f .     Calculate  the  height  of  the  tower. 

12.  Calculate  the  height  of  a  chimney,  if  it  is  found  that  on  walking 
100  ft.  in  a  direct  line  toward  the  chimney  the  angle  of  its  elevation 
changes  from  30°  to  45°. 

13.  Use  tangents  to  calculate  the  distance  apart  of  two  objects  which 
lie  in  a  horizontal  line,  when  it  is  given  that  the  angles  of  depression  of 
the  two  objects  from  a  point  200  ft.  directly  above  the  line  joining  them 
are  45°  and  30°. 

14.  If  the  shadow  of  a  tower  is  60  ft.  longer  when  the  elevation  of 
the  sun  is  30°  than  when  the  elevation  is  45°,  show  that  the  height  of 
the  tower  is  30  (1+V3). 

15.  If  a  stick  27  ft.  long  casts  a  shadow  31  ft.  long,  what  is  the  tan- 
gent of  the  sun's  elevation? 

16.  To  determine  the  distance  between  two  objects,  A  and  B,  on 
opposite  sides  of  a  river,  a  line  100  ft.  long  from  B  to  C  is  measured, 
BC  being  at  right  angles  to  BA.  Calculate  the  lengths  of  AB,  when  the 
angle  BCA  is  30°,  45°,  60°.  Also  when  this  angle  is  one  whose  tangent 
is  1.34. 

17.  If  the  breadth  of  a  house  is  47  ft.,  what  is  the  height,  of  the 
ridgepole  above  the  top  of  the  wall  for  a  plain  roof  when  the  angle  of  the 
roof  is  30°  ?  45°  ?  60°  ?    What  when  it  is  an  angle  whose  tangent  is  0.42  ? 


§  169]  THE   TANGENT   FAMILY.  307 

18.  Given  that  the  ridgepole  of  a  roof  is  23  ft.  above  the  line  joining 
opposite  walls  of  a  house,  what  is  the  width  of  the  house  when  the  slope 
of  the  roof  is  30°  ?  45°  ?  60°  ?  What  when  the  tangent  of  the  angle  of 
slope  is  2.3  ? 

19.  Express  the  length  of  the  side  of  a  regular  polygon  of  n  sides  in 
terms  of  the  radius  of  the  inscribed  circle.  Also  the  radius  of  the 
inscribed  circle  in  terms  of  the  side  of  the  polygon.  What  are  the  cor- 
responding expressions  if  the  circumscribed  circle  is  used  instead  of  the 
inscribed  circle  ? 

20.  From  a  certain  point  the  angle  of  elevation  of  a  kite  is  A.  From 
another  point  due  west  of  the  first  and  a  feet  from  it,  the  kite's  eleva- 
tion is  B.  If  the  first  point  is  due  south  of  the  kite,  express  the  height 
of  the  kite  in  terms  of  A,  B,  a. 

21.  Two  lines,  AB,  AC,  in  the  same  horizontal  plane  and  at  right 
angles  to  each  other,  are  observed  from  a  point  vertically  above  A.  The 
angle  subtended  by  AC  is  one  whose  tangent  is  3;  that  subtended  by 
AB  has  2  for  its  tangent.     If  AC  is  150,  how  long  is  AB! 

22.  Express  the  area  of  a  regular  polygon  in  terms  of  its  side  and  half 
the  angle  subtended  at  its  centre  by  a  side.  Also  in  terms  of  the  radius 
of  the  inscribed  circle.  Also  in  terms  of  the  radius  of  the  circumscribed 
circle. 

23.  If  a  vertical  tower  subtends  an  angle  A  at  a  distance  b  feet  from 
its  foot,  on  the  horizontal  line  through  the  foot,  how  high  is  the  tower  ? 

24.  The  summit  of  a  spire  is  vertically  over  the  middle  point  of  a 
horizontal  square  enclosure  whose  side  is  of  length  a  feet.  The  height  of 
the  spire  is  h  feet  above  the  level  of  the  square.  If  the  shadow  of  the 
spire  just  reaches  a  corner  of  the  square  when  the  sun  has  an  altitude  A, 
prove  that  *  ,    ,- 

25.  Two  stations  due  south  of  a  tower,  which  leans  toward  the  north, 
are  at  distances  a,  b  from  its  foot.  If  A ,  B  are  the  angles  of  elevation  of 
the  top  of  the  tower  from  these  stations,  show  that  the  inclination  of  the 
tower  to  the  horizontal  is  one  whose  tangent  is 

(b  —  a)  tan  A  tan  B 
b  tan  B  —  a  tan  A 

26.  A  pyramid  of  altitude  h  stands  on  an  equilateral  triangle  as  base ; 
express  the  side  of  the  base  in  terms  of  h  and  the  angle  which  the  faces 
of  the  pyramid  make  with  the  base.  If  the  altitude  is  100  ft.  and  the 
apex-angle  is  60°,  show  that  the  side  is  50 VQ. 


308  PLANE   TRIGONOMETRY.  [§169 

27.  A  flagstaff  6  ft.  high  stands  on  the  top  of  a  pyramid  of  square 
base.  When  the  sun's  angular  altitude  is  A,  the  end  of  the  shadow  of 
the  flagstaff  just  reaches  the  edge  of  the  base  of  the  pyramid,  and  at  a 
point  distant  y,  x  from  the  ends  of  a  side  of  the  base.  Show  that  the 
height  of  the  pyramid  is 


4 


t±jl.t-<mA  -6. 


LABORATORY  EXERCISE. 

Pick  out  from  among  the  preceding  exercises  a  few  that  might  possi- 
bly be  of  use  in  practical  life,  and  make  in  the  field  the  measurements 
called  for. 

EXERCISES.-CALCULATIONS   WITH    THE   TANGENT,    USING    THE 

TABLES. 
(Make  diagrams  to  scale  and  test  by  measurement.) 

Note. — The  tangent  of  an  angle  is  looked  up  in  the  tables  in  the  same 
way  as  the  sine,  or  else  by  §  170,  to  avoid  subtraction  of  minutes  and 
seconds. 

Calculated  results  should  show  the  same  number  of  figures  as  data. 
With  data  to  one,  two,  three,  four  figures,  tangents  are  to  be  taken  to 
one,  two,  three,  four  figures  respectively. 

1.  If  the  abscissa  and  ordinate  are  7,  9,  calculate  the  tangent  to  one 
decimal  place  and  determine  the  nearest  angle. 

2.  If  the  abscissa  and  ordinate  are  13,  17,  calculate  the  tangent  to 
two  decimal  places,  and  determine  the  nearest  angle. 

3.  If  the  abscissa  and  ordinate  are  31.4,  29.8,  calculate  the  tangent 
to  three  figures,  and  determine  the  nearest  angle. 

4.  If  the  abscissa  and  ordinate  are  13.45,  23.61,  calculate  to  four 
decimal  places  the  tangent,  and  determine  the  nearest  angle. 

5.  Solve  Exs.  1,  2,  3,  and  4  by  using  logarithms,  and  compare  results 
with  those  just  obtained  by  natural  tangents. 

6.  If  the  abscissa  is  3  and  the  angle  15°,  calculate  to  one  significant 
figure  the  ordinate.     Is  it  best  to  use  logarithms? 

7.  If  the  abscissa  is  4.7  and  the  angle  is  43°,  calculate  by  tangent  to 
two  figures  the  ordinate  to  two  figures.     Is  it  best  to  use  logarithms? 

8.  If  the  abscissa  is  37.2  and  the  angle  is  17°  15',  use  the  tangent  to 
three  figures  to  calculate  the  ordinate  to  three  figures.  Work  the  same 
also  by  logarithms. 

9.  If  the  abscissa  is  43.27  and  the  angle  is  34°  17',  calculate  the  ordi- 
nate to  four  figures,  by  logarithms  and  by  natural  tangents,  and  compare 
results. 


§  169]  THE   TANGENT   FAMILY.  309 

10.  If  the  ordinate  is  7  and  the  angle  55°,  calculate  the  abscissa  to  a 
single  figure  by  the  best  plan. 

11.  If  the  ordinate  is  73  and  the  angle  is  34°  30',  calculate  to  two 
places,  by  the  best  method,  the  corresponding  abscissa. 

12.  If  the  ordinate  is  65.3  and  the  angle  is  54°  15',  calculate  by  loga- 
rithms and  by  natural  tangents  the  abscissa  to  three  figures. 

13.  If  the  ordinate  is  43.76  and  the  angle  is  34°  23',  calculate  to  four 
figures,  by  logarithms  and  by  natural  tangents,  the  abscissa. 

14.  If  the  abscissa  is  —  23  and  the  angle  137°,  calculate  the  ordinate 
to  two  figures. 

15.  If  the  ordinate  is  —  23.4  and  the  angle  200°  15',  calculate  the 
abscissa  to  three  figures. 

16.  If  the  abscissa  is  34.53  and  the  angle  is  1715°  13',  calculate  to  four 
figures  the  ordinate. 

17.  If  the  ordinate  is  5.23  and  the  angle  760°  15',  calculate  the  abscissa 
to  three  figures. 

18.  If  the  abscissa  is  —  4321  and  the  angle  is  —  205°  23',  calculate  the 
ordinate  to  four  figures. 

19.  If  the  abscissa  and  ordinate  are  —  23.5,  54.2,  calculate  the  principal 
angle  to  the  nearest  five  minutes,  and  give  the  general  formula  for  the 
angles. 

20.  If  the  abscissa  and  ordinate  are  —  54.32,  —  67.54,  calculate  the 
principal  angle  to  the  nearest  minute,  and  the  general  angle. 

21.  If  the  abscissa  and  ordinate  are  23,  —  34,  calculate  the  principal 
angle  to  the  nearest  half-degree,  and  the  general  angle. 

22.  If  the  abscissa  and  ordinate  are  7,  -  9,  calculate  the  principal 
angle  to  the  nearest  five  degrees,  and  the  general  angle. 

23.  Calculate  the  area  of  the  regular  polygon  of  eighteen  sides,  if  the 
length  of  a  side  is  34.51. 

24.  A  rectangular  field  has  its  sides  42.14,  67.53.  What  angle  does 
its  diagonal  make  with  the  shorter  side  ?    What  with  the  longer  side  ? 

25.  A  vertical  pole  subtends  an  angle  of  43°  15'  from  a  point  34.5  ft. 
distant  in  a  horizontal  line  from  its  foot.  Calculate  to  three  figures  the 
height  of  the  pole. 


310 


PLANE   TRIGONOMETRY. 


[§170 


§  170.     The  Cotangent. 
The  cotangent  of  an  angle  is  the  reciprocal  of  its  tangent. 

1         abscissa 


p 

y 

„*_ 

S 

' 

^<t 

-o- 

^.Jn 

M 

i 

cot<9  = 


tan# 
cot  0  = 


ordinate 
OM 


Fig.  153. 


MP 

By  §  118,  or  directly  from  a 
diagram,  it  is  clear  that  the 
tangent  of  angle  is  the  cotan- 
gent of  its  complement,  or, 

tan  A  =  cot  (90° -A); 
cot  A  =  tan  (90°  -A); 
By  §  162, 
cot  (180°  +  A)  =  cot  A  ;  cot  (n  ■  180°  +  A)  =  cot  4, 
By  §  163, 

cot  (180°  -  A)  m  -  cot  -A ;  cot  (n  •  180°  -  A)  =  -  cot  ^, 
By  §  108,  tan  B  =  -  cot  (B  -  90°) ;  cot  B  =  -  tan  {B  -  90°), 
tan  C  =  tan  (C  -  180°);  cot  C  =  cot  (C  -  180°), 
tanZ>  =  -  cot  (2>  -  270°);  cotl>  =  - tan  (2>  -  270°). 


EXERCISE. 

1.  Show  that  the  tangent  of  any  angle  of  any  size  can  be  expressed 
as  the  tangent  or  else  the  cotangent  of  an  angle  less  than  45°.  Use  the 
last  six  formulas  for  practice  in  taking  tans  and  cotans  and  their  logs 
from  the  tables,  taking  B,  C,  D  in  the  second,  third,  fourth  quadrant, 
respectively. 

§  171.    Calculations  with  the  Cotangent. 

By  the  definition,  division  by  the  tangent  is  multiplication 
by  the  cotangent,  and  vice  versa. 

(1)  Cotangent  equals  abscissa  divided  by  ordinate. 

(2)  Abscissa  equals  ordinate  multiplied  by  cotangent. 

(3)  Ordinate  equals  abscissa  divided  by  cotangent. 


§  171]  THE   TANGENT   FAMILY  311 

EXERCISES. 

Use  the  set  on  pages  308,  309,  replacing  the  word  "tangent"  by 
"cotangent."  Cotangents  are  looked  up  in  the  tables  by  the  last  six 
formulas  of  §  170. 

THEORETIC  EXAMPLES  ON  THE  COTANGENT. 

1.  A  BM  is  a  straight  line  at  right  angles  to  the  straight  line  MP. 
If  AB  is  of  length  a,  express  MP,  BM  in  terms  of  a  and  the  cotangents 
of  angles  MBP,  MAP. 

2.  Make  up  a  problem  (numerical)  corresponding  to  the  preceding, 
involving  a  church  steeple  on  one  side  of  a  river,  and  two  points,  at 
a  known  distance  apart,  on  the  other  side  of  the  river,  and  calculate  the 
height  of  the  steeple  and  width  of  the  river.  Make  a  diagram  to  scale, 
and  test  by  measurement. 

3.  ABC  is  a  horizontal  triangle,  right  angled  at  C.  AP  is  a  vertical 
line.  CB  is  of  length  a.  Angles  A CP,  ABP  are  <f>,  6.  Express  A P  in 
terms  of  a,  cf>,  0. 

4.  Make  up  a  kite  problem  (numerical)  corresponding  to  Ex.  3,  and 
solve  it,  with  a  diagram  to  scale.     Test  by  measurement. 

5.  Express  the  radius  of  the  circle  inscribed  in  a  regular  polygon  of 
n  sides  in  terms  of  the  side  of  the  polygon  and  half  the  central  angle 
in  circular  measure. 

6.  If  the  side  of  the  polygon  in  Ex.  5  is  34.52  ft.,  calculate  to  four 
figures  the  radius  of  the  inscribed  circle  when  the  number  of  sides  is 
twenty.     Test  with  a  diagram  to  scale. 

7.  Express  the  area  of  a  regular  polygon  of  n  sides  in  terms  of  its 
side  and  half  the  central  angle  in  circular  measure. 

8.  Calculate  the  area  of  the  regular  polygon  of  eighteen  sides,  if 
each  side  is  4.32  ft. 

9.  The  shadows  of  two  vertical  walls,  which  are  at  right  angles  to 
each  other,  and  are  av  a2  feet  in  height,  are  observed,  when  the  sun  is  due 
south,  to  be  bv  b2  feet  broad.  Show  that  if  a  be  the  sun's  altitude  above 
the  horizon,  and  /8  the  inclination  of  the  first  wall  to  the  meridian, 

cot*  a  =  l-\  +  %  and  cot/3  =  ^ 

ttl         a2  a<P\ 

10.  Apply  Ex.  9  to  a  numerical  example.     Can  logarithms  be  used  ? 

11.  Two  stations  due  south  of  a  leaning  tower  (leaning  toward  the 
north)  are  at  distances  a,  b  from  its  foot.  If  a,  /3  are  the  elevations  at 
the  top  of  the  tower  from  these  stations,  show  that  the  inclination  of  the 
tower  to  the  vertical  is  an  angle  whose  cotangent  is 

b  cot  a  —  a  cot  B 
b-a 


312  PLANE  TRIGONOMETRY. 

12.   Apply  Ex.  11  to  a  numerical  problem.     Can  logarithms  be  used? 
Make  diagram  to  scale ;  test  by  measurement. 

EXAMPLES  FROM    RELATED  STUDIES. 

1.  Show  that  if  a  straight  line  passes  through  the  origin  and  is  not 
upright,  the  coordinates  of  any  point  on  it  are  connected  by  the  equation 
y  =  tan  $  •  x,  where  0  is  the  angle  of  the  line  with  the  right-hand  z-axis. 

2.  Show  that  if  the  preceding  line,  instead  of  passing  through  the 
origin,  cuts  off  the  distance  b  from  the  ?/-axis, 

y  =±  tan  0  •  x  +  b. 

3.  Show  that  if  (x,  y)  is  any  point  on  a  straight  line  and  (a/,  y'), 
(x",  y")  are  two  fixed  points,  the  tangent  of  the  angle  which  the  line 
makes  with  the  #-axis  is  any  one  of  the  expressions 


y-ti 

y  -  y" 

x-x"' 

y'  -  y" 

x  —  x' 

x'  -  x" 

y'  -y 

y"  -y 

y"  -  y' 

or 

X'  —  X       X"   —  X       X"  —  X' 

4.  A  particle  has  impressed  on  it  instantaneously  a  vertical  velocity 
of  50  ft.  per  second,  and  at  the  same  instant  a  horizontal  velocity  of 
75  ft.  per  second.  What  is  the  direction  and  what  the  magnitude  of  the 
resultant  ? 

5.  A  cylindrical  stone  tower  20  ft.  high  and  4  ft.  in  diameter  is  tilted, 
by  settling  of  the  earth,  until  just  ready  to  tumble  over.  What  is  the 
tilt? 

6.  A  cube  is  held  on  an  inclined  plane  by  friction.  To  what  angle 
can  the  plane  be  tilted  before  the  cube  rolls  over,  when  the  cube  is  held 
in  the  position  giving  the  easiest  solution  ? 

7.  Solve  Ex.  6,  replacing  the  cube  by  an  upright  cylinder  and  by  a 
parallelopipedon  of  given  dimensions. 

8.  A  body  weighing  100  lb.  is  suspended  by  a  chain.  What  horizontal 
force  will  pull  it  10°  from  the  vertical,  and  what  will  then  be  the  pull 
on  the  chain? 

9.  PB,  PC  are  two  equal  rods,  held  in  a  vertical  plane,  hinged  at  P, 
with  B,  C  in  a  horizontal  line,  B  being  tied  to  C  by  a  string,  and  P  being 
above  the  line  BC.  A  weight  is  set  at  P.  What  pull  will  this  give  the 
string  ?    Solve  some  numerical  examples. 

10.  If  (x,  y)  is  a  point  at  a  distance  b  from  the  lower  end  of  a  straight 
beam  which  slides  in  a  vertical  plane,  one  end  being  on  a  horizontal  plane 
and  the  other  on  a  vertical  plane,  show  that  x  =  a  cos  cf>,  y  =  b  sin  <£, 
where  a  is  the  distance  of  (x,  y)  from  the  upper  end  of  the  beam,  and  <f> 
the  angle  the  beam  makes  with  a  horizontal  iine.     When  is  x2  +  y2  =  a2? 


CHAPTER   IX. 

SINES   AND   COSINES,    TANGENTS   AND   COTANGENTS. 

§  172.    Relation  of  the  Tangent  and  Cotangent  to  Sine  and  Cosine. 

ordinate 

ordinate      modulus        sine 

tangent  =  — = — -. =  — ; — : 

abscissa      abcisssa 


cosine 


modulus 

sin^t 
cos^l 


..cotA=c™A. 
sin  A 

And  since 

sin2  A  +  cos2  ^4  =  1, 

sin2  A      -j  _       1 

COS2  .4.    '              COS2Jl 

d 

1      cos2  A  _      1 
sin2  A      sin2  A 

(4)  is 

1  +  Uui2  A  =  sec2  A. 

(5)  is 

1  +  cot2  A  =  cosec2  A 

...tan  ^  =  ^4-  (1) 


(2) 
(») 

(4) 


(5) 

(6) 
(7) 

.•.tan2^  =  sec2^l-l,  (8) 

and  cot2  A  =  cosec2  A-l.  (9) 

From  (3)  the  sine  can  be  calculated  when  the  cosine  is 
given,  or  the  cosine  can  be  calculated  when  the  sine  is  given. 
From  (6)  the  secant  (or  cosine)  when  the  tangent  is  given. 
From  (7)  the  cosecant  (or  sine)  when  the  cotangent  is  given. 
From  (8)  the  tangent  when  the  secant  (or  cosine)  is  given. 
From  (9)  the  cotangent  when  the  cosecant  (or  sine)  is  given. 

313 


314  PLANE   TRIGONOMETRY.  [§172 

Note  that  since  squares  appear  in  (3),  (4),  (6),  (7),  (8), 
(9),  the  calculated  function  has  two  opposite  values  for  a 
given  value  of  the  given  function.  Explain  this  on  a 
diagram. 

EXERCISES. 

1.  From  the  sine  of  17°  23'  in  the  tables,  calculate  the  cosine  and  com- 
pare with  the  table  value. 

2.  From  the  cosine  of  43°  21'  in  the  tables,  calculate  the  sine  and 
compare  with  the  table. 

3.  From  the  sine  of  15°  32'  in  the  table,  calculate  the  tangent  through 
the  cosecant. 

4.  From  the  cos  67°  41',  calculate  the  tangent  through  the  secant. 

5.  From  the  tan  75°  12',  calculate  the  cosine  through  the  secant. 

6.  From  the  cot  345°  54',  calculate  the  sine  through  the  cosecant. 

7.  From  sin  23°  in  the  table,  calculate  all  its  other  functions  and  test 
such  as  the  tables  give.     From  tan  25°  calculate  the  other  functions. 


§  173.    Use  of  a  Diagram  in  Calculating  One  Function  from 

Another. 

The  formulas  of  §  172  are  used  in  calculating  one  function 
from  another  when  the  given  function  is  in  the  form  of  a  deci- 
mal of  two  or  more  figures.  When  the  given  function  is  in  the 
form  of  a  common  fraction,  it  is  best  to  construct  a  diagram 
corresponding  to  the  given  function,  and  then  read  the  other 
functions  from  the  diagram.  This  is  particularly  true  if  the 
parts  of  the  given  function  are  integers  which  with  some 
other  integer  form  a  right-angled  triangle.  (See  the  set  of 
such  figures  in  the  table  on  page  145.) 

Example.     Given  sin  A  =  f ,  calculate  all  other  functions. 
Figure  154  shows  the  two  positions  of  the  terminal  for 
sin  A  =  | ,  and  gives  ±  4  as  the  third  side. 
And  by  the  diagram : 


§174]     SINES,  COSINES  — TANGENTS,  COTANGENTS.        315 


cosec  A  =  f ; 
tan^.  =  ±f ; 
cot  A  =  ±  f ; 
cos  J.  =  ±  | ; 


sec  J.  =  ±  | ; 
versin  A  —  \  —  cos  .A  =  J  or  -| ; 
►versin  A.  =  1  —  sin  A.  =  |. 


EXERCISES. 


Plot  the  dual  position  for  the  ter- 
minal corresponding  to  each  of  the 
following,  and  calculate  the  ungiven  functions: 


I 

\ 

5 

5 

> 

\ 

:>  . 

S 

yr 

V 

? 

/ 

\ 

/ 

.1 

« 

1.  cos  A  =  fg. 

2.  cos  A  =  -  & 

3.  sin  ^4  =  f $. 

4.  sin  4  =  -  £f 

13.  versin  A  =  ffi. 

14.  versin  A  —  f|f . 


5.  tan  A  =  |f. 

6.  tan  A  =  -  ||. 

7.  cot  4  =  |f 

8.  cot  A  =  -  ff 


Fig.  154. 

9.   cosec  A  =  ffo. 

10.  cosec  A  =  -  | f . 

11.  sec  ^4  =  ff. 

12.  sec  A  =  -  J$l. 

15.  coversin  A  =  T97. 

16.  coversin  A  =  |f . 

17.  Use  the  diagram  for  some  similar  examples  when  the  numbers 
are  selected  at  random,  without  reference  to  making  a  right-angled  tri- 
angle whose  sides  are  integers. 

§  174.     Use  of  a  Diagram  in  Expressing  All  the  Functions 
in  Terms  of  Some  One  Function. 

(a)  All  in  terms  of  the  sine. 

Let  8  =  sine. 

Lay  out  the  double  terminals  where 
s  represents  the  sine  to  the  modulus  1. 
Then  the  abscissa  is  ±  Vl  —  s2. 

From  the  diagram  (Fig.  155)  one 
has  directly  from  the  definitions  of  the 
functions  : 


tan  0  = 


vrr 


esc  0  =  - ;    sec  6  = 
s 


cot  6  = 


±V1-*2 


±V1 

coversin  0  =  1  —  s. 


versin  0  =  1  ±  Vl  —  s2 


316 


PLANE   TRIGONOMETRY. 


[§174 


EXERCISES. 

1.   The  diagrams  shown  in  Figs.  156-160  indicate,  in  order,  given, 
(1)  cosine,  (2)  secant,  (3)  tangent,  (4)  cotangent,  (5)  cosecant.    The 


Fig.  156. — Cosine  given. 


Fig.  157.  —  Secant  given. 


student  may  express,  as  above  with  the  sine,  all  functions  in  terms  of 
each  as  given,  using  the  diagrams. 


Fig.  158.  — Tangent  given. 


— 1 


Fig.  159.  —  Cotangent  given. 


My 


-V5*- I 


2.  Deduce  the  expressions  found  in  the 
preceding  exercise  and  those  in  terms  of  the 
Vo*— 4,  sine,  all  directly  from  the  formulas  of  §  172, 

without  the  use  of  a  diagram. 


Fig.  160.  —  Cosecant  given. 

3.  Show  by  a  diagram  that  sin-1 

4.  Show  by  a  diagram  that  sin-1 


+  VI  +  x2 

1 


=  tan-1  x. 
*=  cot-1  X. 


+  VI  +  z2 
5.  Show  by  a  diagram  that  cos-1 =  cot"1  (x)  =  tan-1  ( - ) . 


6.  Show  by  a  diagram  that  tan-1 


=  sin-1  x. 


+  Vl  -  x2 

7.  What  changes  are  made  in  Exs.  3, 4,  5, 6,  if  the  radical  is  allowed 
the  negative  sign  ? 

8.  If  6  =  sec-1 x  find  from  a  diagram  the  values  of  the  other 

functions.  ±vx  —y 


§174]     SINES,  COSINES  — TANGENTS,  COTANGENTS.        317 


9.   If  6  =  sin-1 ,  find  from  a  diagram  the  values  of  the 


other  functions.      ±  vy*  +  x 


x 


10.   If  6  =  cos-1 ,  find  from  a  diagram  the  values  of  the  other 

functions.  Vy  +x 


x 


11.  If  6  =  tan-1 —  .  find  from  a  diagram  the  values  of  the  other 
functions.                 ^V  ~x 

12.  If  A  =  sin-1 -5 — £=,  find,  in  the  best  way,  the  other  functions. 

x*  +  y*'  j> 

13.  If  sin  A  =       m2  +  2mn        show,  in  the  best  way,  that 

m2  +  2wn  +  2n2'  J' 

.       ,    ra2  +  2  wn 
tan  ^4.  =  ± 


2  mn  +  2  n2 


14.   Prove  sec  (tan"1  a;)  =  ±vl  +  a;2. 


15.  Prove  sin  (cos-1  x)  =  ±  Vl  —  a:2. 

16.  Prove  cos  (cot-1  a;) 


17.   Prove  cot  (sin-1  x) 


±Vl  +  a:2 
±VT^x* 


x 

18.  Select  the  cosine  of  some  angle  in  the  table  of  cosines  and  calcu- 
late the  other  functions  and  compare  results  with  the  table. 

19.  Do  the  same  as  indicated  in  Ex.  18  for  each  function  given  in  the 
table,  selecting  for  each  a  different  angle. 

EXERCISES  CONNECTING  TANGENT  AND  COTANGENT  WITH 
THE  OTHER  FUNCTIONS. 

1.  tan  A  +  cot  A  —  sec  A  cosec  A. 

2.  (tan  A  +  cot  ^4)2  =  sec2  A  +  cosec2  A. 

3.  l  +  tan^==sin^+cos^t 

sec  A 


4.  (1  +  tan  ^4)(1  +  cot  A)  =  2  +  sec  A  cosec  A. 

-  cosec  A  nnB  A 

5. =  cos  A. 

cot  A  +  tan  A 

-  1  —  tan  A      cot  .4  —  1 
o.  = • 

1  +  tan  A      cot  -4+1 

7  1  +  tan2  A  _  sin2  A  < 
1  +  cot^-cos2^" 


318  PLANE   TRIGONOMETRY.  [§175 

8.  =sin^4  cos  .4. 

cot  A  +  tan  A 

9.  cos8  A  tan  A  +  sin2  A  cos  A  tan  A  =  sin  A. 

sec  A 


10.   (tan  A  —  sin  A  )  cosec8  J. 


1  +  cos  ^4 


,,     cot  J.  —  tan  A  _  cos  ^4  +  sin  A 
cos  J.  —  sin  A        cos  -4  sin  A 


12.  V!  ~  S^n  A  =  sec  yl  -  tan  .4. 

*  1  -f  am  A 

13.   - -=  sec  .4  +  tan  ^4. 

sec  A  —  tan  A 

14.  sec  A  -  tan  A  =  j  __  2  sec  4  tan  ii  +  2  tan2 ,4. 
sec  ^4  +  tan  A 

,c       tan  J.       .      cot  ^4  A  ,1,1 

15. +  ; =  sec^l  cosec A  +1. 

1  —  cot  A      1  —  tan  A 

n  c        cos  A       ,      sin  A  A    ,  . 

16. + =  sin  A  +  cos  A . 

1  —  tan  A      1  —  cot  A 

17.  (sin  A  +  cos  A)  (cot  A  +  tan  A)  =  sec  ^4  +  cosec  A. 

18.  cot4  ^4  +  cotM  =  cosec4  A  —  cosec2  A. 

19.  sec2  J.  cosec2  A  -  tan2  ^4  +  cot2  A  +  2. 

20.  tan2  A  -  sin2  J.  =  sin4  J.  sec2  ^4. 

21       cot  A  cos  ^4    _  cot  A  —  cos  ^4 
cot  A  +  cos  J.        cot  A  cos  -4 

22.  cot  j  +  tan  g=  cot  A  tang. 

cot  i*  +  tan  A 

23    tan  ^  +  sec  A  —  1  _  1  -f  sin  A  # 
tan  A  —  sec  A  +  1         cos  J. 

24.   j  ~  s?n  4  =  1  +  2  tan  ,4  (tan  4  -  sec  A). 

1  +  smA 

25.   (tan4+cosec£)2-(cot5-sec^)2=2tan^cot£(cosec4  +  secJB). 


§  175.  The  Addition-subtraction  Formula  for  Tangents. 

tan(,l  +  i?)=sin(A  +  i?)  (1) 

V  }     cos(A  +  £)         •  W 

_  sin  J.  cos  B  +  cos  J.  sin  B  w 

cos  A  cos  ^  —  sin  .4  sin  1? 


§175]       SINES,  COSINES  — TANGENTS,  COTANGENTS.        319 

Dividing  numerator  and  denominator  by  cos  A  cos  B,  this 

becomes 

sin^i      sin  B 

s  a       x»n        cos^l      cos  B 
tan  (A  +  B)  = — — — —  (3) 

1  _  sin^-  sin  -" 

cosJL  cosi? 

or,  ten{A  +  B)=te\A+^B.  (4) 

1  -  tan  A  tan  B  v  J 


Similarly, 

an  f  A  _   Tt\  —  _ 

1  +  tan  ^  tan  B' 


Ni-B)=^^  (5) 


cot(^  +  B)=cot^cot^-1>  (6) 

cot-B  +  cot^fi  v 

cot  (4  -  B)  =  gotBcot^  +  l  (7n 

k  ;      cotJS-cot^l  w 


EXERCISES. 

1.  From  the  values  of  tan  45°  and  tan  30°  show  that  tan  75°  =  2  +  V3 
=  3.73205-... 

2.  Similarly,  tan  15°  =  2  -  V3  =  0.26795-.. . 

3.  Tan  R  =  \,  tan  B  =  f.     Find  tan  (R  +  B),  tan  (i2- 5),  cot  (R  +  B), 
cot  (R-B),  tan  (2  22  +  £),  tan  (2  #  -  B). 

4.  Take  D,  Z?  as  some  angles  in  the  tables,  and  calculate  tan  (D  +  B), 
tan  (D  —  B),  cot  (D  +-B),  cot  (D  —  B),  and  compare  results  with  the  tables. 

5.  *»n  *  ~  ten  ?  =  ten  (s-y)  tans, 
cot  x  +  tan  y 

6.  Prove  tan  2  x  =    2tan*    ,  and  find  tan  (A  +  B  +  C). 

1  —  tan2  x 

7.  Prove  cot  2  x  =  cot*x~19  and  find  value  of  cot  3  x. 

2  cot  a: 

8.  If  tan  L  —  \  and  tan  B  =  |,  find 

tan  (£  +  5),  tan  (L  -  JB) ;  cot  (L  +  5) ;  cot  (L  -  B). 

9.  Show  by  formula  (5)  that  tan  0°  =  0,  and  by  (4)  that  tan  90°  =  <x> . 


320  PLANE   TRIGONOMETRY.  [§176 

§  176.    The  Addition-subtraction  Formula  in  Inverse 
Tangents. 

If  tan  A  =  x  and  tan  B  =  y, 

A  +  B  =  tan"1  x  4-  tan"1  y. 

Thus  (4),  (5),  of  the  preceding  article  may  be  written  in 
the  forms 

tan-1a;  +  tan-12/  =  tan-1^^.  (3') 

"  1+xy  v    J 

Similarly,  if  cot  A  =  x  and  cot  B  =  y  (6),  (7),  take  the 
forms :  t 

cot-1aj  +  cot-12/  =  cot1^-^.  (6') 

x  +  y  K    J 

cot-1*;  -  cot-1^  =  cot1^^.  (7') 

Re-read  what  has  been  said  about  multiplicity  of  values  of  sin-1  x, 
cos-1  x. 

EXERCISES. 

1.  tan"1  \  +  tan-1  T^  =  tan"1 1. 

2.  tan-1  £  +  tan-1  i  =  mr  +  j- 

3.  t«^*_  *»-*•=*  «,,,  +  ». 

rc  m  +  n  4 

4.  tan-1 1  +  tan-1 -  =  tan_1- — -• 

1  - 12  1  -  3  *2 

5.  tan-1 1  +  tan-1  f  s  |  cos-1  f . 

6.  cos"1 1  +  tan-1  f  =  tan"1  § f 

7.  2  cos-1——  +  cot-1 1|+ \  cos-1  ^  has  7r  for  one  of  its  values.   Give 

V13 
also  the  general  value.     Test  formulas  above  by  tables. 

8.  sin-1 — -  +  cot-13   has  j  for  one  value.    Give  also  the  general 
value.  ^5  4 

9.  tan-1  f  +  tan-1 1  -  tan"1  T\  has  |  for  one  value. 

10.  If  tan-1 1  +  tan-1  x  =  tan-1  f ,  find  the  general  value  of  x. 

11.  tan-1 1  +  tan-1 }  +  tan"1 1  +  tan"1  \  =  J  (one  value). 


§177]     SINES,   COSINES  — TANGENTS,   COTANGENTS.        321 

p  —  ox  p 

12.  tan-1 1,1,  =  tan"1  -  -  tan"1  x. 

q  +  px  q 

p  -\-  ox  ,  p 

13.  tan-1  - — i-  =  tan"1  L  +  tan"1  x. 

q  -px  q^ 

14.  tan-1  — ^—  =  tan-1  a;?  —  tan-1  a;. 

l  +  *i 

1 
x  — 

15.  cos-1 J  =  2oot-1x.    (Set  a;  =  cot  0.) 

a: 

16.  sin-1 -^- =  2  tan-1  x.     (Set  a:  =  tan  0.) 

1  +  x2  v 

17.  sec-1  ^    }     •  s  2  cos-1  a:.     (Set  a;  s  cos  0.) 

2  a:2  —  1  v 


§  177.    Tangent  of  the  Double  Angle. 

From  tan  (4  +  2>)  .   tan  ^  + tan  ^ 

1  —  tan  .A  tan  i? 

follows,  if  A  =  B, 

2  tan  ^ 


tan  2  ^1  = 


1-tuPA 


Thus  the  tangent  of  any  angle  is  twice  the  tangent  of  half  of 
it  divided  by  one  minus  the  square  of  the  tangent  of  half  of  it. 


2tan| 

.*.  tan  x  = ; 

l-tan*f 

tan8*=2tanf;f    ; 
1- tan2  4  a; 

to^.J^SJg    ,  etc. 
1  —  tan2  2  x 


LABORATORY    EXERCISE. 


On  a  ten-inch  circle,  measure  tan  5°  and  tan  10°,  and  test  the  pre- 
ceding formula. 


322  PLANE   TRIGONOMETRY.  [§  177 

EXERCISES. 

1.  Test  tan  42°  18'  in  the  tables  as  compared  with  tan  21°  9'. 

2.  If  tan  A  =  §,  find  tan  2  A,  tan  3  A,  and  tan  4:  A. 

3  tan  A  -  tan3  A 


3.  Show  tan  3  A  = 


1-3  tan2  ;1 


4.  Show  tan  4  A  =    H^A-t^A) 

l-6tanM  +  tan44 

sin  2  ^4 

5.  Deduce  tan  2  ^4  from  tan  2  4  = -— -  by  using  the  formulas  for 

cos  2 ,4     J         h 
sin  2  A  and  cos  2  A  in  terms  of  functions  of  A. 

A  «    -r.  A  Jl  —  cos^l 


1  _  tan2  -  8#   Prove  tan  ^  =  ±  yj- 


6.  Prove  cos  ^  = 4-  2  '  1+COB  * 

1  +  tan2-  ^_l-cos^ 

9.   Prove  tan2  -      sinA 

2  tan  - 
„    ™  .-i  2  ,,*t^  ,      i        sin  i 

7.  Prove  smi= j-  10.   Prove  tan  -~  = j 

1  +  tan2-  J      2  cos2-- 

2  2 

11.  Prove  that  tan  —  always  has  the  same  sign  as  sin  A.    Can  this  be 
shown  on  a  diagram  ? 

12.   Prove     sin^    =cot-- 
versin  A  2 


14    -o  •    ^4         /versing 

13.  Prove  sin  —  =  -vl 

2*2 

14.  Prove  cotAA  +]  =  sec2A  +  tan 2 ,4. 

cot  A  —  1 

^4  yl 

15.  Prove  tan  —  +  cot  —  =  2  cosec  ^4. 

—  — 

16.  Prove  cot  A  -tan  v4  =2  cot  2  A. 

17.  Prove  2  cosec  2  A—  tan  ^4  +  cot  A. 

18.  Prove     2  C0JA    =  tan  2 ,4. 

cotM-1 

19.  Prove  cosec  2  A  +  cot  2  ^4  =  cot  ^4. 

20.  Prove  cos^  +  sil^  _  cos  ,4- sin  ^  =  2  tan  2  At 

cos  A  —  sin  ^4      cos  A  +  sin  yl 

21.  Prove  (1  +  tan  ,4)  cot  .4 


1  —  tan  A 


§178]     SINES,   COSINES  — TANGENTS,   COTANGENTS.        323 

22.  Prove  i  +  i«»'(4y-^)  m    i    =  cosec 2 ^ 

1 -tan2(45°-^l)      sin2A 

23.  1±8in2^  =  tan(450±^). 

cos  2  A 

24.  tan  (45°  +  2  J  )  +  tan  (45°  -24)  =  2sec44. 

25.  If  tan2  x  =  1,  what  is  the  general  value  of  a;? 

§  178.   Tangent  of  the  Donble  Angle  in  Inverse  Notation. 

From  tan  2  A  =  -  2  tan  f   . 

l-tan2^l 

follows,  if  tan  A  =  x  and  .A  =  tan-1  a?, 

2  /■ 

or,  2  tan -*x  =  tan _ * 5 •     Test  by  tables. 

1-a?2 

EXERCISES. 

1.  2  tan-1 1  =  tan-1  V  (one  value). 

2.  2  tan"1  £  +  tan"1 }  +  2  tan"1  J  =  J  (one  value). 

3.  4  tan-1 1  -  tan-1  J*  +  tan"1  JU  =  ^  (one  value). 

4 

4.  3  tan-1  J  +  tan"1  ^  =  J  -  tan"1  T^J5  (one  value). 


5.  tan'1  Vl  +  x*  ~  1  =  £  tan-*  x. 

*  +  i 

a? 

6.  sec-1 =-  =  2  cot-1 2r. 

a; 

x 

7.  cos-1  (2  x1  -  1)  =  2  cos"1  x. 

8.  sin-1-?-^-  =  2tan-1a:. 

1+z2 

9.  sec-1 =  2  cos-1  x. 

2x2-  1 

10.  cos-1  5— —  =2tan"1a;. 
1  +  z2 


324  PLANE   TRIGONOMETRY.  [§  179 


11.  tan"1  p^^  =  S  tan"1  x. 

12.  sin-1  (3  a; -4  a:8)  =  3  tan-1 


y/l-x* 


13.  cos-1  (4  xs  -  3  x)  =  3  tan 

x 


14.   sin"1 2  a;  VI  -  x2  =  2  tan"1 


vT^iT2 


15.  cos-1  (2  x2  -  1)  =  2  cot"1 


VI  -x* 


16.   cos"1  (1-2  a;2)  =  2  tan-1 


VI  -x2 


§  179.     Tangent  of  the  Half-angle. 

'2tan^ 
Since  tan  A  = 


1  _  tan24' 
2 

if  we  let  tan  A  =  a  and  tan  —  =  x,  x  is  gotten  by  solving  the 
quadratic 

ax2  -f-  2  x  —  a  =  0. 


#  = 


a 


.   ^M^_-l=bVl  +  tan2^ 

.  .  i«iii  -oj-  — t 5 • 

2  tan  J 

If  tan  A  is  given,  there  are  two  positions  of  the  correspond- 
ing terminal,  180°  apart.  There  are  thus  two  positions  of 
the  terminal  of  the  half-angle,  90°  apart.  This  explains  the 
dual  sign.  When  A  is  itself  given  and  not  by  its  tangent, 
the  proper  sign  is  to  be  selected.     Thus 

tan22°=-1+V:L  +  tan2-i5 
tan  45° 

and         tan  100°  =  =  1  ~  Vl  + ta"2  20±°- 

tan  200° 


§179]      SINES,   COSINES —  TANGENTS,   COTANGENTS.        325 

The  tangent  of  the  half-angle  can  also  be  expressed  in  terms 
of  other  functions. 

It  has  been  proven  (§  139)  that 

1-  cos  A=2  sin2 4,  (1) 

l  +  cos^l  =  2cos2^,  (2) 

A       A 

and  sin  A  =  2  sin  —  cos  —  •  (3) 

—         — 


-WJ-^Si  w 


■••Mi 

To  get  (4),  divide  (1)  by  (2);  to  get  (5),  divide  (1)  by 
(3);  to  get  (6),  divide  (3)  by  (2).  The  dual  sign  in  (4) 
means  that  if  an  angle  is  given  by  its  cosine,  there  are  two 
positions  of  the  terminal  (as  in  §  106).  Thus  there  are  two 
positions  for  the  half-terminal,  and  two  values  of  the  tangent 
of  the  half-angle,  oppositely  equal.  When  the  angle  itself 
is  given,  there  is  only  one  terminal  and  the  proper  sign  is  to 
be  selected  in  (4).     Thus 


tan  11°  =  +Vj-7C0SoL  and  tan  100°  =  -  Vt"" 
1 1  +  cos  22  Tl  + 


-  cos  200° 


cos  200c 


EXERCISES. 

1.  Express  tan  22^°  in  radicals.     Express  the  radicals  in  decimals  and 
calculate  tan  22^°,  and  compare  with  tables.   Do  the  same  for  tan  (  —  22^°). 

2.  Express  tan  15°  in  radicals  and  in  decimals,  and  compare  with 
tables. 

3.  Express  tan  7\°  in   radicals  and  in  decimals,  and  compare  with 

tables.     Test  the  four  formulas  for  tan  —  by  tables. 

A  ^ 

4.  If  cos  A  =  .23,  find  tan  — ,  and  explain  the  dual  sign. 

A  ^  A 

5.  If  sec  A  — 1\,  find  tan  —  and  tan  A.    Express  —  in  inverse  tangents. 


326  PLANE  TRIGONOMETRY.  [§  180 

6.  tan  ( 45°  +  ^.)=J]  + sin  A=sec  A +  ta,nA. 

\  2  /      *  1  -  sin  A 

7.  sec/|  +  $\ sec (|  -  $\  =  2  sec  2  ft 

8.  If  sin  $+smcf>=a,  and  cos  0  +  cos  <f>  =  b,  find  the  value  of  tan  :— x. 

9.  If  sin  J.  =  ■§,  find  sin  2  A,  cos  2  .4,  tan  2  ^4,  and  tan  — 

m 

10.  If  cos  ^4  =Hj  nn<l  sin  2  ^4,  cos  2  ^4,  tan  2  ^4,  and  tan  —  • 

11.  If  tan  A  =£§,  find  tan  2  ^4,  sin  2  A ,  and  cos  2  A. 

12.  If  sec  .4  =  f,  find  all  the  functions  of  2  A. 

13.  If  cosec  J.  =  \S  find  all  the  functions  of  2  ^4. 

1-tan2- 
2 

14.  ^-  =  cos^4. 

1  +  tan2^ 

-4  ^4 

15.  1  -f  cot  ^4  cot  —  =  cosec  A  cot  —  • 

-  _ 


16 


.ta„(|  +  |)ta„(|-*)=l. 

17.  tan  -  +  2  sin2  -  cot  a:  =  sin  x. 

18.  tanf  J  +  -)  =  sec  a:  +  tan  a:. 

19.  tan2|(l  +  cot2^V=4cosec2^  =  4(1  +  cot2 ,4). 

§  180.    The  Tangent  in  Auxiliary  Formulas. 

When   forms  like  a  cos  x  -f  b  sin  x  (§  148)  are  to  be  cal- 
culated, and  logarithms  are  to  be  used,  we  write 

a  cos  x  +  b  sin  x  =  Va2  +  &2( —  a        cos  a:  -t sin  x\ 

Wa*  +  b*  Va2  +  62  / 

We  may  set  —  a        =  sin  6  ;   —     .       =  cos  <f>, 

where  tan  </>  =  -  is  more  readily  calculated  than  sin  </>  or  cos  <£. 

.-.  aco8x  +  bsmx=  Va2  +  52 sin  (x  +  <£) 

=  Va2  +  62  sin  fa  +  tan"1 |Y 


§180]     SINES,  COSINES  — TANGENTS,  COTANGENTS.        327 

EXERCISES. 

1.  Find  x  when  317  sin  x  +  212  cos  x  =  321. 

2.  Find  x  when  8.314  sin  x  -  7.215  cos  a;  =  1.314. 

3.  Show  VB  •  sin  x  +  cos  z  =  2  sin  (  x  +  -  ] . 

4.  Show  3  sin  x  +  4  cos  a;  =  5  sin  f  a;  +  tan-1-  J. 

5.  Express  the  preceding  results  in  terms  of  the  cosine  and  cotangent. 

6.  Express  a  cos  x  —  b  sin  x  in  terms  of  the  sine  and  in  terms  of  the 
cosine. 

7.  Solve  the  examples  under  §  148  when  the  tangent  expresses  the 
auxiliary  angle. 


EXERCISES  CONNECTING  THE  TANGENT  AND  COTANGENT  WITH 
OTHER  FUNCTIONS. 

x    sin  A  +  sin  B  =  ^  A±B  ^  ^-^ 
sin  yl  —  sin  2?  2  2 

2  cosyl+cosi?  m  CQt  ^+£  cQt  A-B 
cos  i?  —  cos  A  2  2 

3  8inil+BinB  =  tanii+g 
cos  A  +  cos  5  2 

4  sin  yl  —  sin  Z?  _      .  A  +  B 
cos  5  —  cos  A  2 

-    sin  7  ^4  —  sin  5 .4      ...  A 

5-  s- ; — -  =  tan4. 

cos  7  ^4  +  cos  5  ^4 

6.  sin^+sin3^=tan2^t 
cos  ^4  +  cos  3  A 

7.  cos2^  +  cos2^  ^      5)  ^  _  ^ 
sin2i*-cos2,4            v            y        v  ' 

Q    sin  yl  +  sin  2  A  .A 

8. — -=cot— • 

cos  A  —  cos  2  ^4  2 

9.  sin5-4-sin3-4  =  tan^. 
cos  3  A  +  cos  5  ^4 

10    sin2^  +  sin2^  =  ta  ^ 

sin  2 ,4  -  sin  2  £  v  y       v  y 

lx    cos21?-cos2,4  =        (4  _  B) 
sin  2  5  +  sin  2  4  v  ' 


328  PLANE  TRIGONOMETRY.  [§  180 

12.  si°(^4-2i?)  +  sin(4£-2,l)  A 
cos(4:A-2B)  +  008(45  -2  A)           K  ' 

,  0     tan  5A  +  tan  3 .4       .        n  4        A  A 

13.   — !- =  4  cos  2 .4  cos  4  ^4. 

tan  5  A  —  tan  3  A 

14  cos3^+2cos5^  +  cos7^=cos2^_sin2^tan3^- 

cos  A  +  2  cos  3  v4  -f  cos  5  ^4 

15  s*n  -^  4-  sin  3  ^4  +  sin  5  J.  +  sin  7  ^4  _  .       ** 
cos  yl  +  cos  3  A  +  cos  5  yl  +  cos  7  ^4 

16  s*n  fj  —  sin  5  ^4  +  sin  9  A  —  sin  13  A  _      .  *   * 
cos  A  —  cos  5  A  —  cos  9  A  +  cos  13  ^4 

sin  2  ^ 


1  +  cos  2  ^4 

18. 

sin  2  A             ,   . 

=  cot  A. 

1  —  cos  2  A 

19. 

1  —  cos  2  A           0  A 

— —  =  tan2  A 

1  +  cos  2  A 

20. 

tan  A  +  cot^4  =  2cosec2yt. 

21. 

tan  A  —  oot  A  =  —  2  cot  2  A. 

22. 

cosec  2  A  +  cot  2  ^4  =  cot  A. 

23. 

°osA  -.=**(&  ±4) 

1  T  sin  A            \           2  / 

24. 

sec8^-1-tan8^-cot2, 

sec  4  ^4  —  1 

0(-     1 +  tan2(45°- ^4)  0  , 

25.  — ! -v. -f  =  cosec  2^1. 

1-  tan2(45°-.4) 

26.  If  sin  0  +  sin  <£  =  a,  and  cos  0  +  cos  <f>  =  b,  find  0  and  </>. 

27.  tan(45°  +  4)  -  tan(45°  -  A)  =  2tan2.4. 

28>   cos.4  ±  sin  A  _cosA  -  sin  A  =  2tan2^t 
cos  J.  —  sin  .4      cos  ^4  +  sin  ^4 

29.  cot  (JL  +  15°)  -  tan  (A  -  15°)  =     4  cos  2  ^ 


1  +  2  sin  2  A 

sin  yl  —  cos  A 
1  +  cos  J.  +  cos  2  ^4  ~"  1  +  sin  A  +  cos  A  ~    '"  2 


30.       sin^+sin2^      =tKaA.        31.   *  ±  sin^  ~  cos^  =  tan 


§180]     SINES,   COSINES  — TANGENTS,   COTANGENTS.        329 


32  sin  (w  + 1)4  -  sin  (n- 1)4  _        A 

'   cos(n  +  l)4  +  2cosn4  +  cos(n  -  1)4         D~2* 

„    sin  (n  +  1)  A  +2  sin  nA  +  sin  (n  —  1)  A  A 

33.      2 — — \        1 =   COt-pr- 

cos  (n  —  1)  A  —  cos  (n  +  1)  A  2 

34.  cot-  -tan-  =  2  cot  A. 

2  2 

35.  cot  4  +  cot  (60°  +  A  )  -  cot  (60°  -  4)  =  3  cot  3  4. 

36.  tan  3  A  tan  2  4  tan .4  =  tan  3  4  -  tan  2 4  -  tan4. 


2  tan: 


1  -  tan2- 


37.  sin  A  = 


38.   cos  A  = 


1  +  tan2: 


1  +  tan2|. 


39.  tan  6°  tan  42°  tan  66°  tan  78°  =  1. 

40.  Given  sin  (A  —  x)  =  cos  {A  +  x),  find  tan  x. 

41.  Given  sin  (x  +  a)  +  cos  (x  +  a)  —  sin  (x — a)  +  cos  (x — a) ,  find  tan  x. 

42.  Given  tan  A  tan  a:  =  tan2  (A  +  x)  —  tan2  (4  —  x),  find  cos  a\ 


43.  Given 
Prove 

44.  Given 
Show  that 


m  tan  {A  —  x)  _       n  tan  x 
cos2  x  cos2  (-4  —  x) 

tan  (4  -  2  x)  =  — -  tan  A. 


tan  (4  +  x)  tan  (4  —  a:)  = 


n  +  m 

1  -2  cos  2  4 

l  +  2cos24* 

a:  =  30°  is  a  solution. 


Show 


45.   Given     n  sec2  x  tan  (4  —  x)  =  m  sec2  (4  —  x)  tan  a\ 

n  sin  2  4 


46.   Prove 


tan  2  a: 
1  +  sin  A 


1  —  sin  A 


m  +  n  cos  2  4 
tan2f45°  +  ^V 


47. 


tan  (45°  +  —\  +  cot  ^45°  +  ^\  =  2  sec  4 . 


48. 

tan 

(30° 

+  4)  tan 

(30°- 

1x_2cos2^4  -1 
;     2cos24+l 

49. 

If 

sin  a;  = 

sin4 

sin  (B  +  x), 

show 

tanx  = 

sin 

4  sin  B 

1  —  sin  4  cos  B 
50.   If  m  sin  B  cos  (4  —  x)  —  n  sin  4  cos  (B  +  ar),  find  tana;. 


330  PLANE   TRIGONOMETRY.  [§  181 

§  181.    Trigonometric  Equations  involving  All  the  Functions. 

Process  for  Solution :  Reduce  the  equation  to  a  single  func- 
tion and  solve  in  the  most  general  manner,  using  tables  if 
necessary. 

Sample  Example  : 

sin  x  —  tan2  x,  (1) 

sin2  x  ^0x 

.-.  sma?  = — —  •  (2) 

cos^a; 

•••  (1  —  sin2  x)  sin  x  =  sin2  x.  (3) 

.\  sin  x  (1  —  sin2  x  —  sin  x)  =  0.  (4) 

.%  sin  x  =  0 ;  (5) 

or,                              1  —  sin2  x  —  sin  x  =  0.  (6) 

By  (5),  x  —  iwr,  or  n  •  180°. 

By  (6),         sin x  =  ~1^V5  =  0.6180,  or  - (1.6180). 

St 

The  latter  is  rejected,  since  sin  x  lies  between  +1,  —1. 
If  x  =  sin"1  0.6180; 

x=2n  180°  +  38°  10'  (tables); 
or,  x  =  (2  n  +  1)  180°  -  38°  10'. 

EXERCISES. 
Solve  in  the  most  general  manner : 

1.  sec2  x  =  4  tan  x.  3.  2  cot2  0  =  cosec2  0. 

2.  4  tan  x  —  cot  a:  =  3.  4.  4  cos  0  —  3  sec  0  =  2  tan  0. 

5.  sin2  0  -  2  cos  0  +  \  =  0. 

6.  cot2  0  +  (\/3  +  ^)  cot  0  +  1  =  0. 

7.  tan2  0  +  cot2  0=2. 

8.  sec0-  l  =  (\/2-l)tan0. 

9.  tan20-(l+V3)tan0+V3  =  O. 
10.  2cos0  =  V3cot0. 

11.  1.4  tan  0  +  1.7  cot  0  =  2.3.  14.  tan2  0  +  cot2  0  =  *£- 

12.  2134  sin  0  =  1372  tan  0.  15.   tan2  a:  +  cosec2  x  =  3. 

13.  43  tan  0  =  27  sin2  0.  16.  2  sin2  0  +  3cos0  =  0. 


CHAPTER  X. 


THE  TANGENT   AND   COTANGENT   IN   THE   SOLUTION 
OF   OBLIQUE-ANGLED   TRIANGLES. 

§  182.  Given  Two  Sides  and  the  Included  Angle. 

Since  for  a  solution  by  sines  a  pair  of  opposites  must  be  given, 
if  two  sides  and  the  included  angle  are  given,  the  triangle  can- 
not be  solved  by  sines  without  splitting  the  triangle  into  two 
right-angled  triangles.     A  solution  by  the  aid  of  the  tan- 


gent is  possible.         Giyen 


b,  c,  A. 

sin  C      c  vs      J 


Assume  b>  c. 

sin  B  —  sin  C  _  b  —  c 
sin  B  +  sin  0      b  +  c 


sin 


B-0         B  +  C 


cos 


by  §  145, 


COS 


B 


or, 


tan 


B-0 


C     .B+C 

_ .  sm  __ 

cot*±^ 


b-c, 
b  +  c9 

b-c 
b  +  c 


Since 


2  b  +  c  2 

A  +  B+  (7=180°, 

-L>    ~\-      C    QQO    ^l 

2  2* 

B+C 


(*) 


tan 


=  tan  (90°-^=  cot 


(The  tangent  of  an  angle  is  the  cotangent  of  the  comple- 


ment.    §  170.) 


tan 


B-C 


b  +  c       2 

331 


(D 


332 


PLANE   TRIGONOMETRY. 


[§182 


From  formula  (X),  or  from  (Y),  by  logarithms,  or  by 

B  —  O  B  4-  0 

natural  functions,  — - —  can  be  found.     And  since  — ^— 

is  known, 

B=B+C     B-0 


and 


0= 


2  2     7 

B  +  0     B-0 


2  2 

B  and  O  being  thus  known,  there  remains  to  find  a. 

b 


or, 


a  =  sin  A 


a  =  sin  A 


sini?' 


"  sinC 

In  a  calculation,  the  values  of  B,  O  should  be  checked. 
For  this  purpose  there  may  be  used  b  •  sin  0  =  c  ■  sin  B. 

LOGARITHMS. 

b=  =e 

sin  (7  =  =  sin  B 


sum 


sum 


Check  to  within  2  in  "  final "  figure. 

Since  a  can  be  found  by  two  sine  formulas,  that  one  not 
used  for  calculating  a  may  be  used  as  a  check  on  a. 

Model  Example. 


Given 


r  6  =  48.3 
J  <?  =  36.0 


tan 


B-  0     b 


Find 


e.      B  +  O 

tan 


2 
B-0 

2 
B 
0 
a 


=  13°  40' 

=  72°  40' 
=  45°  20' 
=  44.7 


2  b  +  c  2 

6-  <?=  12.3;         5  +  <?=  84.3; 


a  =  sin  A  • 


sin  (7' 


182] 


TRIANGLES  SOLVED  BY  TANGENTS. 


333 


log  tan  =  log  (&-<?)+  log  tan  — ± log  (&  +  <?); 

log  a  =  log  sin  ^1  +  log  c  —  log  sin  (7. 

Logarithmic  Check  for  5,  C. 

LOGARITHMS. 


tan 


LOGARITHMS. 

5 -<?=  1.0899 

B+C 


0.2212 
2 

sum  =  1.3111 
b  +  <?  =  1.9258 
diff.  =  1.3853 


6  =  1.6839 
sin  (7=1.8520 


1.5563=  (7 
1.9798  =  sin.£ 


This  gives 


log  tan 
3-  C 


B-0 


to  the 


nearest  5  minutes  as  13°  40' 

LOGARITHMS. 

sin^L  =  1.9459 

c  =  1.5563 

sum  =  1.5022 

sin  (7=1.8520 
a  =1.6502 


1.5359   1.5361 

The  data  in  lines  being 
three-figured  and  the  check 
holding  to  within  2  in  third 
place,  B,  C  are  assumed 
correct. 


Logarithmic  Check  for  a. 


a  =  1.6503 
sin  £=1.9798 


1.6839=6 
1.9459  =  sin  A 


1.6301    1.6298 

Agree     to    three     figures 
within  2.     Satisfactory. 


EXERCISES. 

Prove  directly  the  formulas  corresponding  to  that  proven  when  ft,  c, 
A  were  given,  when  there  are  given: 

1.  a,  b,'  C,  with  a  >  b.  4.   6,  c,  A,  with  c  <  b. 

2.  a,  b,  C,  with  a  <  b.  5.   a,  c,  B,  with  a  >  c. 

3.  b,  c,  A,  with  c  >  b.  6.   a,  c,  Z?,  with  a  <  c. 

7.  What  if  the  two  given  sides  are  equal? 

8.  The  student  may  construct  some  appropriate  examples,  solve  and 
test  them  (§  77)  : 

(a)  With  lines  to  one  significant  figure  and  angles  reading  to  the 
nearest  5°. 

(b)  With  lines  to  two  significant  figures  and  angles  reading  to  the 
nearest  half  degree. 


334  PLANE   TRIGONOMETRY.  [§  183 

(c)  With  lines  to  three  significant  figures  and  angles  reading  to  the 
nearest  five  minutes. 

(d)  With  lines  showing  four  significant  figures  and  angles  reading  to 
minutes. 

(e)  With  lines  showing  five  significant  figures  and  angles  reading  to 
seconds. 

(/)  Six-figured  data ;  angles  to  tenths  of  seconds. 

(g)  Seven -figured  data ;  angles  to  hundredths  of  a  second. 

9.   Show  that  a  can  also  be  calculated  from 

C  +  B  .     C  +  B 

cos  — — —  sin  ~£ — 

«  =  (*  +  &) _!—=(<._&) 


C-B      v         '   .     C-B 

cos sin  


§  183.  To  find  the  Angles  of  a  Triangle  by  Tangents  when  the 
Three  Sides  are  Given. 

0.  sin  a; 

bmce  tan  x  = , 

cos  a; 

we  have,  from  the  values  of  the  sines  of  the  half  angles  of  a 
triangle  in  §  92  and  the  values  of  the  cosines  in  §  127,  the  fol- 
lowing values  of  the  tangents  of  the  half  angles  of  a  triangle  : 


2       *     s^s  -  a) 

(1) 

2       '       s(s-b) 

(2) 

(3) 

with  their  reciprocals  for  the  cotangents  of  the  half  angles. 
For   calculation  purposes   it  is  best  to  multiply  (1)  by 

under  the  radical ;   (2)  by y ;   (3)  by  — — — 

Then,  tan  *  =  -±-  .aBSO=JL; 
As  —  a     ^  s  s —  a 


:?-J_  JO  -  «X* -&)(«- <0  _    n 

2      s-b'v  s  "s-b9 


tan 


2  ~*_*    M 


)(s -£)(*-<?)         r, 


«  —  <?     *  s  s  —  c 


§  183]  TRIANGLES   SOLVED  BY   TANGENTS.  335 

Calculation  Scheme. 


a= 

NUMBERS. 

(1) 

b  = 

(2) 

c= 

(3) 

2  8  = 

(4) 

8  = 

(5) 

s  —  a= 

(6) 

8-b  = 

(7)     Check 

8  —  C  = 

(8)     (6) +  (7) +  (8)  =  (5) 

LOGARITHMS. 

s  —  a  = 

(1) 

8-b  = 

(2) 

8  —   C  = 

(3) 

sum  = 

(5) 

*= 

(4) 

(5) -(4) 

r?  = 

(6) 

(6)+2 

U  = 

(7) 

a)-a) 

tan|= 

(8) 

a> -(2) 

tan  — = 
2 

(9) 

(7) -(3) 

tan-= 
2 

(10) 

The  scheme  is  made  before  beginning  the  work,  then  numbers 
are  entered  on  the  scheme.  The  logarithms  (1),  (2),  (3), 
(4)  are  all  looked  up  before  their  manipulation  begins  in 
(5),  etc.  The  numbers  are  close  enough  on  the  scheme  to 
be  added,  subtracted,  without  rewriting.  If  not,  the  log  (7) 
may  he  written  on  the  end  of  a  card  and  held  in  turn  over 
(1),  (2),  (3),  and  the  subtractions  be  made  and  written  in 
their  appropriate  place. 

Since  the  tangent  solution  requires  but  four  logarithms, 
those  of  s,  s—  a,  s  —  b,  s  —  c,  it  is  more  convenient  for  use  than 


336  PLANE  TRIGONOMETRY.  [§  183 

the  corresponding  formulas  for  sines  and  cosines,  which 
required  six,  seven,  logarithms.     (See  §§  92,  127.) 

As  the  tangent,  over  much  of  the  tables,  varies  more 
rapidly  than  the  sine  and  cosine  for  equal  changes  in  the 
angle  (notice  this  in  the  tables),  the  tangent  formulas  are, 
as  a  rule,  more  suitable  for  calculation  than  the  cosine  and 
sine  formulas.  This  is  again  taken  up  in  Chapter  XI.  Be- 
sides, the  log  tan  will  give  the  angle  when  it  is  near  90°  as 
well  as  when  small,  even  though  the  tangent  is  infinite. 

Checks. 
A  +  B  +  C=180°  (1) 

or  a  •  sin  2?  =  b  •  sin  A  and  a  •  sin  C=  c  •  sin  J.  (2) 

How  close  the  second  check  should  check  we  have  often  had 
occasion  to  point  out.  How  close  the  first  check  should  check 
depends  likewise  on  the  number  of  significant  figures  in  the 
data,  if  they  are  assumed  as  measurements  and  not  as  exact. 

When  the  sides  are  given  to  one,  or  two,  or  even  three, 
significant  figures,  and  represent  measurements,  the  summa- 
tion of  angles  is  not  an  appropriate  test.  Better,  then,  test 
by  logarithms  of  the  sine  formulas  (second  test  above). 

It  is  often  said  that  if  a  five-place  table  is  used  the  test 
should  hold  to  within  "a  few  seconds."  And  so  it  should, 
for  exact  data.  That  is  absurd,  however,  on  measurements, 
for  no  one  of  the  angles  may  be  knowable  to  minutes.  In 
one-figured  data  (measurement)  the  angles  cannot  be  known 
even  to  the  nearest  degree. 

It  cannot  be  impressed  upon  the  student  too  often  that 
there  is  no  such  measurable  triangle  as  one  having  its  sides 
as  9,  8,  7.  The  more  accurately  one  tries  in  a  physical  labo- 
ratory to  measure  the  length  of  a  given  line  the  more  will 
the  results  differ.  One  can  measure  a  line  roughly  and  get 
the  same  result  each  time.  But  if  the  same  line  is  measured 
with  extreme  care  each  time,  the  results  rarely  agree.  It  is 
quite  absurd,  therefore,  to  say,  "if  five-placed  tables  are 
used,  the  angles  should  sum  to  180°  to  within  a  few  sec- 


§183]  TRIANGLES  SOLVED  BY   TANGENTS.  337 

onds."  It  is  childish  to  use  a  five-place  table  on  one-figured 
data  to  the  full  extent  of  the  table.  When  a  table  of  more 
places  than  the  data  is  used,  calculated  results  should  be  cut 
back  to  the  pattern  of  the  data.  The  severity  of  a  check 
must  in  all  cases  correspond  to  the  character  of  the  data. 

If  one  is  using  a  five-place  table,  with  data  appropriate  to 
such  a  table,  it  is  easy  to  test  the  closeness  with  which  the 
summation  of  the  angles  should  reach  180°.  In  §§  22,  24, 
it  is  shown  that  a  logarithmic  formula  like  that  of  the  tan- 
gent of  the  half-angle  of  a  triangle  may  be  in  error,  on 
account  of  the  approximate  values  of  the  tangents  in  the 
tables,  by  about  1  in  the  last  place  of  logarithms.  How 
much  this  means  for  the  angle  will  depend  upon  the  size  of 
the  angle.  The  possible  error  for  each  angle  may  be  deter- 
mined by  examining  the  table  in  its  neighborhood.  Take  the 
sum  for  the  three  angles  and  double  it  and  we  have  the 
allowable  error  in  the  check. 

Consider,  for  example,  an  equilateral  triangle.  The  half- 
angle  is  30°.  In  the  neighborhood  of  30°  in  a  four-place 
table  a  difference  of  0.0001  in  logarithms  of  tangents  means 
a  difference  of  about  20"  in  angles.  So  in  such  a  triangle, 
calculated  from  its  sides  by  tangents  of  half-angles,  the  sum 
of  the  three  angles  may  differ  from  180°  by  2f.  The  use  of 
a  five-place  table  on  data  appropriate  for  the  use  of  such  a 
table  to  five  places,  will  cut  this  by  the  divisor  10  on  each 
angle,  making  the  allowable  variation  in  the  check  about 
12  seconds. 

How  close  the  check  should  check  has  to  be  determined  for 
each  triangle. 

As  a  general  rule  holds  for  the  logarithmic  check,  it  is  best  to 
use  that. 

To  say,  however,  that  the  place  of  the  table  used  deter- 
mines how  close  the  check  should  check,  without  reference 
to  the  character  of  the  data,  is  to  assume  the  triangle  being 
dealt  with  a  theoretic  triangle.  Such  triangles  are  of  no 
value  to  the  engineer. 


338  PLANE   TRIGONOMETRY.  [§  183 


Model  Example. 

NUMBERS. 

a  =  15.47 

b  =  17.39 

c=  22.88 

2s  =  55.74 

s  =  27.87 

s  -  a  =  12.40 

s-b  =  10.48 

r-  c=    4.99 

LOGARITHMS. 

s-a=  1.0934 

s-5  =  1.0204 

s-  c  =  0.6981 

&m  =  2.8119 

s  =  1.4451 

r«a=  1.3668 

r{  =  0.6834 

tan  4  =1.5900 
2 

.-.  4=21°15'(+) 

tan  |=1.6630 

.-.  -  =  24°  43' 

2 

tan -  =  1.9853 

2 

.-.  !Z  =  44°    2' 
Z 

^  =   42°  31' 

.  B  «    49°  26' 

(7=    88°    4' 
180°    1' 
At  21°  15',  the  log  difference  0.0001  means  about  15". 

At  24°  43',  the  log  difference  0.0001  means  about  20". 

90" 
At  44°    2',  the  log  difference  0.0001  means  about  ^—. 

55  " 
So  the  solution  may  be  considered  correct. 

It  is  clear  that  the  angles  may  sum  to  180°  and  still  the 
individual  angles  be  incorrect.     This  is  rather  unlikely. 

The  use  of  a  five-place  table  with  the  data  of  the  foregoing 
example  (representing  measurements)  could  in  no  wise  affect 
the  closeness  of  the  check,  since  with  data  to  only  four  sig- 
nificant figures,  a  fifth  figure  in  the  tangents  of  angles  would 


§  183]  TRIANGLES   SOLVED  BY  TANGENTS.  339 

be  without  significance.  However,  if  we  assume  the  data  as 
exact,  representing  a  theoretic  triangle,  a  five-place  table  will 
demand  a  closer  check  than  a  four-place  table.     See  §  77. 

Logarithmic  Test, 
logarithms. 

a  =  1.1895         1.2403  =  6 
nBtm  1.8806         1.8298  =  sin  A 


1.0701         1.0701 
They  agree  too  well !     C  should  also  be  checked. 

EXERCISES. 

1.  The  student  may  select  some  numerical  exercises,  solve  and  test : 

(a)  With  sides  to  one  significant  figure,  calculating  angles  to  the 
nearest  five  degrees. 

(b)  With  two-figured  data ;  angles  to  the  nearest  half  degree.  » 

(c)  With  three-figured  data ;  angles  to  the  nearest  five  minutes. 
(c?)  With  four-figured  data ;  angles  to  the  nearest  minute. 

(e)  With  five-figured  data ;  angles  to  the  nearest  second. 
(/)  With  six-figured  data ;  angles  to  tenths  of  a  second. 
(<7)  With  seven-figured  data ;  angles  to  hundredths  of  a  second. 

2.  In  Fig.  145,  line  DEC  <  DAC  <  BBC.     Show  from  this  that 

b  a 

sin  0  <  0  <  tan  0,  and  - — ^  =  1,  when  0  =  0 ;  also ^  =  1,  when  0  =  0. 

sin  0  tan  0 

3.  In  Fig.  145,  triangle  OEC  <  sector  OAC  <  triangle  OBC.     Show 
from  this  that  sin  0  <  0  <  tan  0. 

4.  Show  that  (1  +  sin  $y°™8  =  e,  when  $  =  0. 

5.  Show  that  (1  +  3  tan2  0)2cosecMe  ^  g^  when  0  =  0. 

6.  From  the  series  for  sin  6  and  cos  0,  find  a  series  for  tan  0. 

7.  Given  r  sin  <f>  =  21.71,  r  cos  <f>  =  31.41,  find  r,  <£. 

8.  Given  r  cos  <£  cos  6  =  3.172,  r  sin  <f>  cos  0  =  2.113,  r  sin  0  =  3.121, 
find  r,  0,  cji. 

9.  Given  3121  tan  <£  +  2171  cot  <£  =  3141,  find  <£. 

10.  Given  sin  (<£  +  16°  17')  =  0.3142  sin  <f>,  find  <£. 

11.  Given  tan  (<£  +  31°)  =  23  tan  <£,  find  <£. 


12.   Change  va2  —  x2  and  va2  +  x2  to  trigonometric  forms. 


CHAPTER   XI. 

GENERAL    REVIEW    ON    THE    SOLUTION    OF    TRIANGLES. 
LIST   OF   FORMULAS   TO   MEMORIZE. 

§  184.     Solution  of  Triangles. 

It  is  possible,  as  we  have  seen,  to  solve  any  sort  of  triangle 
with  the  sine  alone.  Similarly,  we  might  use  the  cosine 
alone,  or  the  tangent  alone.  In  fact,  since  all  the  trigo- 
nometric functions  are  expressible  in  terms  of  any  one  of 
them,  any  one  of  them  might  be  used  alone  for  the  solution 
of  triangles. 

EXERCISES. 

1.  Show  that  it  is  possible  to  solve  all  possible  cases  in  right-angled 

triangles  by  using : 

(1)  The  sine  alone. 

(2)  The  cosine  alone. 

(3)  The  tangent  alone. 

2.  Show  that  for  triangles  not  right-angled  it  is  always  possible,  either 
directly,  or  by  breaking  the  triangle  into  two  right-angled  triangles,  to 

solve  every  case : 

(1)  By  the  sine  alone. 

(2)  By  the  cosine  alone. 

(3)  By  the  tangent  alone. 


§  185.    The  Best  Method  of  Solving  Triangles. 

What  is  the  best  method  in  any  particular  case  involves 
two  considerations :  (a)  reaching  an  answer  with  the  least 
labor,  (/3)  reaching  an  answer  the  most  accurate  possible 
from  the  data.     Evidently,  if  the  observer  is  the  computer, 

340 


§  185]  GENERAL  REVIEW.  341 

he  should  know  beforehand  what  method  of  computation  is 
contemplated  and  avoid  observations  leading  to 

(a)  Angles  near  90°  to  be  computed  from  the  sine. 
(5)  Small  angles  to  be  computed  from  the  cosine. 

EXERCISES. 

1.  Consider  all  possible  data  in  right-angled  triangles,  and  determine 
what  formulas  will  give  answers  with  the  least  labor. 

2.  Do  the  same  with  triangles  not  right-angled. 

3.  In  case  the  three  sides  of  a  triangle  are  given  and  all  three  angles 
are  desired,  how  many  logarithms  are  required  if  (i)  sines  are  used,  (ii)  if 
cosines  are- used,  (iii)  if  tangents  are  used? 

4.  When  is  it  better  to  calculate  by  the  natural  functions  rather  than 
by  logarithms?  Investigate  as  to  whether  practical  engineers  of  your 
acquaintance  use  logarithms  or  natural  functions  in  ordinary  calculations. 

It  is  frequently  stated  that  an  angle  can  be  calculated 
better  from  the  log  tan  than  from  the  log  sin  or  log  cos. 

The  table  on  page  342  will  show  that,  with  a  five-place 
logarithmic  table, 

(i)  Angles  less  than  about  25°  can  be  calculated  as  ac- 
curately from  log  sin  as  from  log  tan. 

(ii)  Angles  between  about  65°  and  90°  can  be  calculated 
as  accurately  from  log  cos  as  from  log  tan. 

(iii)  Angles  between  25°  and  65°  can  be  calculated  more 
accurately  from  the  log  tan  than  from  log  sin  or  log  cos. 

(iv)  An  angle  of  any  size  can  be  calculated  as  accurately 
from  log  tan  as  from  log  sin  or  log  cos. 

(v)  Small  angles  cannot  be  calculated  at  all  accurately 
from  the  log  cos. 

(vi)  Angles  near  90°  cannot  be  calculated  at  all  accurately 
from  log  sin. 

(vii)  Log  tan  (log  cot)  varies  the  same  way  at  both 
ends  of  the  table,  so,  therefore,  when  an  angle  is  near  90°  its 
value  is  given  as  readily  by  log  tan  as  if  it  were  near  0°, 
even  though  tan  90°  =  qo. 


342 


PLANE   TRIGONOMETRY. 


[§185 


A° 

Sin 

Tan 

AND  Cot 

Cos 

A° 

d 

60" 
d 

d 

60" 

d 

d 

60 
d" 

1° 

724 

0".08 

724 

0".08 

0.22 

273" 

89° 

2° 

362 

0.17 

362 

0.17 

0.44 

136 

88° 

3° 

241 

0.25 

242 

0.25 

0.66 

91 

87° 

4° 

181 

0.33 

182 

0.33 

0.89 

67 

86° 

5° 

144 

0.42 

145 

0.41 

1.1 

55 

85° 

10° 

72 

0.8 

74 

0.8 

2 

30 

80° 

15° 

47 

1.3 

50 

1.2 

3 

20 

75° 

20° 

35 

1.7 

40 

1.5 

5 

12 

70° 

25° 

27 

2.2 

33 

1.8 

6 

10 

65° 

30° 

22 

2.7 

29 

2.1 

7 

9 

60° 

35° 

18 

3.3 

27 

2.2 

9 

7 

55° 

40° 

15 

4.0 

26 

2.3 

11 

5 

50° 

45° 

13 

4.6 

25 

2.4 

13 

5 

45° 

60" 
corresponds  to  — - 


The  columns  headed  d  are  the  differences  in  the  logarithms 
for  1'  in  the  neighborhood  of  the  angles  given  at  the  side, 
and  since  sin  A  =  cos  (90  —  A),  and  tan  A  =  cot  (90  —  A),  a 
difference  like  724  for  sine,  opposite  1°,  is  the  difference  for 
cos  89°,  etc.    Since  d  is  for  1',  a  difference  of  1  in  logarithms 

The  table  gives,  at  various  parts  of  a 

five-place  logarithm  table,  the  difference  in  seconds  corre- 
sponding to  a  difference  of  1  in  logarithms. 

In  the  neighborhood  of  1°  a  difference  of  1  in  logarithms 
corresponds  to  a  difference  of  about  0.08"  for  angles  obtained 
from  log  sin  and  for  log  tan,  and  to  about  273"  for  angles 
obtained  from  log  cos. 

Such  an  angle  can,  therefore,  be  determined  with  equal 
accuracy  from  log  sin  and  log  tan,  but  cannot  be  deter- 
mined accurately  from  log  cos,  where  any  interpolation  is 

necessary. 

60" 
Note  that  up  to  about  25°  the  values  of are  nearly  the 

same  for  sine  as  for  tangent  and  cotangent. 

Therefore  angles  less  than  about  25°  can  be  determined 

almost  as  accurately  from  the  log  sin  as  from  log  tan,  and, 


§186]  GENERAL  REVIEW.  343 

consequently,  those  between  65°  and  90°  can  be  determined 
almost  as  accurately  from  the  log  cos  as  from  log  tan. 

It  is  thus  clear  that  the  angle  can  be  determined  at  any 
place  as  accurately  from  the  log  tan  as  from  the  log  sin  or 
log  cos,  while  between  25°  and  65°  it  can  be  determined 
more  accurately  from  the  log  tan  than  from  either  log  sin 
or  log  cos,  since  there  a  difference  of  1  in  logarithms  cor- 
responds to  a  smaller  difference  for  tangents  than  for  sines 
or  cosines. 

In  a  table  of  six,  seven,  ten  places  the  advantage  in  calcu- 
lating from  the  tangent  would  appear  stronger  than  it  does 
here  with  a  five-place  table. 

§  186.    Angles  determined  when  they  are  Small  from  Log  Sin 
and  Log  Tan. 

Large  tables  are  arranged  to  give  such  angles  from  the 

formulas 

log  sin  x  —  log  x"  +  S 

log  tan  x  =  log  x"  +  T. 

See  Gauss's  or  Hussey's  Tables.  The  student  may  be  assigned  some 
exercises. 

EXERCISES. 

1.  Explain  from  the  formula  tan  x  =  ^LE  why  at  the  beginning  of 

COS  X 

the  table  the  difference  for  log  sin  is  about  the  same  as  for  log  tan, 
and  near  90°  the  difference  for  log  tan  is  about  the  same  as  for  log  cos. 

2.  Explain  why  the  difference  for  log  tan  is  everywhere  the  same 
as  for  log  cot. 

3.  Explain  why  the  differences  for  log  sin  decrease  from  0°  to  90°. 

4.  Explain  why  the  differences  for  log  tan  and  log  cot  decrease  from 
0°  to  45°,  where  they  are  least,  and  then  increase  to  90°,  being  the  same 
at  equal  distances  from  0°  and  90°,  and  reconcile  this  with  tan  90°=oo  and 
log  tan  90°  =  go,  and  explain  why  it  is,  in  a  table  running  to  45°  with 
function  and  co-function,  like  Gauss's  or  Hussey's,  an  angle  near  89° 
can  be  determined  as  accurately  from  log  tan  as  an  angle  near  1°. 


344  PLANE   TRIGONOMETRY.  [§186 

WRITTEN    REVIEW  EXERCISE. 

1.  Make  out  a  statement  for  the  possible  data  in  right-angled  tri- 
angles, with  the  formulas  best  adapted  for  calculation  in  each  special 
case,  with  suggestions  as  to  what  should  be  done  in  exceptional  cases, 
when  certain  angles  or  lines  are  small. 

2.  Do  the  same  for  triangles  not  right-angled. 
The  following  general  conclusions  should  appear : 

(a)    In  right-angled  triangles  : 

(i)    The  cosine  (secant)  connects  the  hypothenuse  and 

the  side  bordering  the  angle, 
(ii)    The  sine  (cosecant)  connects  the  hypothenuse  and 

the  side  opposite  the  angle, 
(iii)    The  tangent  (cotangent)  connects  the  two  sides. 

(5)    In  triangles  not  right-angled : 

(i)   When  a  pair  of  opposites  are  given,  use  the  sine. 
(ii)    When  three  sides  are  given,  use  tangents  of  half- 
angles, 
(iii)    When   two   sides    and    the    included    angle    are 
given,  get  the  two  remaining  angles  by  tangents 
(§  172),  and  the  third  side  by  sines. 

3.  Make  a  tabulated  statement  of  the  formulas  and  processes  of  trigo- 
nometry which  should  be  committed  to  memory.  (The  extent  of  this 
will  depend  upon  the  career  in  view  by  the  class,  and  the  outline  is  left 
to  the  teacher.) 

4.  Show  that      c  =  a  cos  2?  +  b  cos  A,  (1),  §  125, 

becomes,  on  making 

a  =  sin  A  — - — ,  b  =  sin  B  •  - — — , 
sin  C  sin  C 

since  sin  C  =  sin  (A  +5), 

sin  (A  +  B)  =  sin  A  cos  B  +  cos  A  sin  B  (§  136). 

Show  that  this  is  a  general  proof. 
Show  also,  from  Fig.  127,  that 

-cosc^+n;-/-62 

2  ah 
.\  cos  (A  +  B)  =  cos  A  cos  B 
Show  that  this  proof  is  general. 


x_y_-t 

h    a     b 

a 

smA  sin 

B. 

CHAPTER  XII. 


THE   QUANTITY    V^~l   IN   TRIGONOMETRY. 


§  187.   The  Argand  Diagram. 


X 

^^~    * 

>^. 

41^ 

N 

/_ 

>s 

t^^ 

$ 

-      ^      ^ 

1 

Q_     . 

0                 "      -A 

L 

_i    _ 

A 

7 

^ 

7 

Ss 

J? 

^^^^ 

r^ 

Fig.  163. 


If  OA,  00  (Fig.  163)  are  equal  in  length  but  opposite  in 
direction, 

0O=-0A  =  (-l)-0A. 

Thus  multiplying  a  line  by 
(—1)  is  equivalent  to  a  turn 
through   180°. 

Let  i  (the  initial  letter  of 
imaginary)    stand    for    V  —  1. 

Then  w  =  -l. 

...  00=(-l)'OA  =  i-i-OA. 

Thus,  multiplying  twice  by  i 
is  equivalent  to  a  turn  through 

180°.     Consequently,  a  reasonable  interpretation  of  multiply- 
ing once  by  i  is  a  turn  through  90°. 

.-.  i-OA=OB, 

i.i-0A=(-l).0A  =  0O, 

i.i'i'OA  =  (i'i)-(i-OA)  =  (-l)-OB=OD, 

i-i-i'i-OA=(i.i)>(i'i).OA  =  (-l)-(-l)-OA  =  OA 

Thus,  if  numbers  expressed  in  terms  of  the  unit  (+1)  are 
laid  out  to  the  right  on  a  horizontal  line,  those  in  terms  of 
the  unit  ( —  1)  are  to  the  left  on  the  same  line,  while  those 
in  terms  of  the  unit  (+ 1)  are  laid  out  on  the  upright  vertical 
and  those  in  terms  of  the  unit  (—  i)  are  on  the  downright 
vertical. 

345 


346 


PLANE   TRIGONOMETRY. 


[§187 


Numbers  of  the  form  a  +  bi,  where  a,  b  are  in  terms  of  the 
units  (1),  (—  1),  one  or  both,  as 

5  +  5t\     («) 

-5  +  5*,     (/3) 

-5-5i,     (7) 

5-5*,    (S) 

are  called  complex  numbers. 

They  are  plotted  by  laying  out  a,  according  to  sign,  from  the 
vertical  axis  along  the  horizontal  axis,  and  then  laying  out  5, 

according  to  sign,  from  the  ter- 
minal of  a,  parallel  to  the  vertical 
axis.  The  points  Pv  P2,  P3,  P4 
(in  Fig.  164)  represent  the  num- 
bers (a),  (/3),  (7),  (8)  above,  a 
side  of  one  of  the  small  squares 
being  taken  as  a  unit. 

When  a,  b  are  given  all  real 
values,  positive,  negative,  ra- 
tional, irrational,  from  zero  to 
infinity,  the  expression  a  -f  bi  is 
a  varying  quantity,  and  is  repre- 
sented, in  a  one  to  one  correspondence,  by  the  points  of  the 
plane,  each  point  representing  a  number  and  each  number 
represented  by  a  point.  Such  a  diagram-representation  of 
a  +  bi  is  called  the  Argand  Diagram,  from  the  French 
mathematician  who  first  made  the  use  of  it  prominent, 
though  priority  of  suggestion  is  accredited  to  Gauss. 


Fig.  164. 


EXERCISES. 

On  coordinate  paper  plot  the  numbers  : 

±1±»;  ±2±4t;  ±3±2i;  ±-L±-Lt;  ±i±^»;  a  +  5»; 

V2      V2  2 

x  +  yi ;  ±  cos  25°  ±  t  sin  25° ;  ±  cos  30°  ±  t  sin  30° ;  ±  cos  100°  ±  •  sin  100° ; 

±  cos  190°  ±  i  sin  190° ;  ±  cos  280°  ±  i  sin  280° ;  2  (cos  30°  +  i  sin  30°)  ; 

3  (cos  120°  +  i  sin  120°)  ;  5  (cos  103°  +  i  sin  103°) 

V2  (cos  315°  +  i  sin  315°)  ;  cos  797°  +  i  sin  797°. 


188] 


THE   FOUR  UNITS. 


347 


§  188.   The  Addition  (Subtraction)  of  Complex  Numbers. 

(a  +  hi)  +  (c  +  di)  =  (a  +  c)  +  (b  +  d>'. 

Pj  is  a  +  6i,  if  Oil^  =  a,  and  Tlf^  =  6,    (Fig.  165) 

and  P2  is  c  -f-  cfa',  if  OM2  =  c,  and  ifcf2P2  =  d. 

If  the  parallelogram  on  OPv  OP2  is  completed  as  in 
Fig.  165,  evidently, 

P3  is  (a  +  c)  +  (b  +  <*>',  with  Oif3  =  a  +  *,  and  MZP3  =  b+d. 

Thus,  to  add  P2  to  P2,  start  with  Pv  and  draw  a  line, 
PXPZ,  parallel  and  equal  to  OP2  (and  in  the  same  direction). 
The  point  reached,  P3,  represents  the  sum,  P1  -f  P2. 


ii 

i 

/ 

> 

r< 

' 

:i 

.- 

-' 

[' 

/ 

•-' 

-' 

, 

/ 

r> 

/ 

r 

/ 

- 

. 

n 

, 

J* 

/ 

- 

I 

- 

^ 

■ 

£. 

j 

( 

) 

v. 

u 

' 

/ 

" 

Fig.  165. 


Fig.  166. 


Similarly,  since  subtraction  is  the  reverse  operation  to 
addition,  to  subtract  P2  from  Pj  (Fig.  166),  start  with  Pv 
and  lay  out  P^s  parallel  and  equal  to  P20  (not  0P2). 

Thus,  points  are  added  (subtracted)  as  forces  in  mechanics. 


EXERCISES. 

The  teacher  may  select  some  examples  for  plotting  additions,  sub- 
tractions, making  the  selection  so  that  pairs  of  points  fall  in  the  same 
quadrant,  and  in  different  quadrants  (all  possible  combinations).  Use 
coordinate  paper  (ten  by  ten  to  the  inch  will  be  found  convenient). 


348 


PLANE   TRIGONOMETRY. 


[§189 


IT 

^*            ^^  T^ 

7                   /vf- 

2                 .1/    \    . 

_r       -    —\r    -*\ 

7              -A          V 

X-7 

n           M , 

I                          *  ■    I 

L                                                                                i 

\                                                                       / 

s               z 

^           ^ 

Fig.  167. 


§  189.     Points  on  the  Unit-circle,  and  on  Any  Circle. 
The  point  Px  in  the  unit-circle  evidently  represents 

cos  6  +  i  sin  6. 

Thus,  if  6  is  given  all  values  from  0 
to  2  7r,  the  expression  cos  6  +  i  sin  0 
will  represent  all  points  on  the  unit- 
circle. 

Similarly,  if  a,  b  vary  subject  to  the 
condition  Va2  +  62  remaining  constant, 
the  expression  a  +  bi  will  represent  all 
points  on  the  circle  of  radius  Va2  -f  b2. 
If  a,  5  are  constants,  a  +  bi  represents  a  point  at  a  distance 

Va2  +  b2  from  the  origin  and  in  the  direction  tan-1  -. 

Thus  a  complex  number  depends 
upon  two  things,  the  distance  and  the 
direction  of  its  representing  point  from 
the  origin.  The  former  is  called  the 
modulus  of  the  number,  also  its  absolute 
value.  The  latter  is  called  the  ampli- 
tude, or  angle.  For  the  amplitude,  the 
smallest  positive  angle  less  than  360° 
locating  the  modulus  is  taken. 

a  +  bi  =  Va2  + 


_  __L 

Z*~          *     71 

/                X°jt^ 

Z                       ^  v 

J~                  ■*'    $ 

~X           ^L  c 

t          ^  $£-*  - 

I                       a .     '     - 

A                               1 

_r                      y 

i                             L 

^           ^ 

5*.        ,*7 

\ 

Fig.  168. 


Va2  + 


f  f 


Va2  +  &2> 


=  Va2  +  b2  (cos  (9  +  *  sin  0). 

Thus,  a  4-  bi  has  two  factors,  a  length  factor  and  a  direct- 
ing factor. 

The  radical,  +  Va2  +  b2  =  r,  is  the  modulus,  or  length 
factor. 

The  factor,  cos  0  +  i  sin  0,  is  the  directing  factor. 

The  angle  0  is  the  amplitude. 

The  radical,  +  Va2  4-  £>2,  is  an  absolute  number,  or  a  num- 
ber in  terms  of  the  unit  (+1). 


§190] 


THE   FOUR   UNITS. 


349 


Since  a  4-  hi  =  (cos  6  4-  i  sin  0)  • 
Va2  +  62,  and  since  the  first  factor  rep- 
resents a  point  on  the  unit-circle,  we 
may  look  upon  the  point  P,  represent- 
ing a  +  bi,  as  derived  from  the  corre- 
sponding point  P1  on  the  unit-circle  by 
multiplication  by  the  absolute  number 
+  Va2  +  b2.     Thus  with 


• 

7                     S\&fr>t 

rf      -e""""5k"    ^    \ 

±  /  zyM^~p 

:M:::^:::: 

-t    s::::z:  j_: 

_^    _:=-s:      zL_ 

mS^:::^::: 

Fig.  169. 


a  +  bi  =  (cos  0  +  i  sin  0)  •  Va2  4-  62 
and  c  +  di  =  (cos  <£  4-  i  sin  <£)  •  Vc2  4-  d2, 

(a  4-  bi)  (c  4-  ^0 


=  {(cos  6  4- » sin 0)(cos  <£  +  t sin  <£)|  •  { Va2  4  62  •  V^  +  d2j. 

Hence  the  multiplication  of  any  two  complex  numbers  can 
be  reduced  to  the  multiplication  of  two  numbers  (points)  on 
the  unit  circle,  together  with  the  multiplication  of  two  abso- 
lute numbers  as  in  arithmetic. 

§  190.    Multiplication  (Division)  of  Points  on  the  Unit-circle. 

(cos  6l  4-  i  sin  0j)(cos  62  4-  i  sin  #2) 
=  (cos  01  cos  02  —  sin  01  sin  #2)  - 

4- 1  (sin  0j  cos  02  4-  cos  01  sin  02) 
=  cos  (^  +  02)  4  i  sin  (0X  4-  02)- 
or,  PxPt  =  Pz. 

Thus  any  two  points  on  the  unit- 
circle  are  multiplied  by  adding  (alge- 
braically') their  angles. 
Since  division  is  the  reverse  operation  of  multiplication, 
P1  on  the  unit-circle  is  divided  by  P2  on  the  same  circle  by 
subtracting  (algebraically)  the  angle  of  P2  from  that  of  Pv 
Similarly,  the  point  representing  the  product  of  three  such 
points  has  its  angle  equal  to  the  algebraic  sum  of  the  angles 
of  the  factors. 

Cor.  :  In  general,  (cos  0  4-  i  sin  0)n  =  cos  n6  +  i  sin  n  0, 
when  n  is  a  positive  integer. 


IE 

-■i*7'--==-,N"S- 

/\       a     t-fl   /L  \    -L 

^               ^ 

:::s-:::::;-z::: 

___S,----,,Z 

Fig.  170. 


350  PLANE   TRIGONOMETRY.  [§  191 

EXERCISES. 

1.  Prove  by  radicals  and  direct  multiplication  that  (cos  30°  +  i  sin  30°)2 
=  cos  60°  +  i  sin  60°,  and  that  (cos  60°  +  i  sin  60°)2  =  cos  120°  +  i  sin  120°. 

2.  Prove  by  dirept  multiplication,  using  the  tables,  that  (cos  10°  -f 
i  sin  10°)  (cos  20°  +  i  sin  20°)  =  cos  30°  +  i  sin  30°. 

3.  Prove  by  using  radicals,  with  direct  multiplication,  that  (cos  30°  + 
i  sin  30°)3  =  i ;  (cos  120°  +  i  sin  120°)3  =  1 ;  (cos  60°  +  i  sin  60°)3  =  -  1 ; 
(cos  300°  +  i  sin  300°) 3  =  -  1. 

4.  Prove  by  using  the  tables  that  (cos 36°  +  /sin  36°) 2  =  cos 72°  + 
i  sin  72°. 

5.  Prove  by  using  radicals,  with  direct  multiplication,  that  (cos  45°+ 
i  sin  45°)*=  -  1 ;  (cos  135°  +  i  sin  135°) 4  =  -  1 ;  (cos  225°  +  i  sin  225°)4  = 
-  1 ;  (cos  315°  +  i  sin  315°)4  =  -  1. 

6.  Use  a  diagram  (Groat's  coordinate  paper  is  convenient)  to  solve 
the  preceding  examples. 

7.  Prove   directly,  and   by  diagram, =  cos  6  +  i  sin  6, 

cos  6  —  i  sin  6 
and  show  that  if  Px-  P%  =  1,  Pv  P2  are  symmetric  to  the  horizontal  axis. 

§  191.   Multiplication  of  Points  not  on  the  Unit-circle. 

This  may  be  carried  out  in  three  ways, 
(a)  Direct  multiplication. 
Example  :  2  +  3  i 

4-5t 

8  +  2i+  15  =  23  +  2i. 

This  method  is  suited  to  the  case  when  a,  b  in  a  +  hi 
are  small. 

(6)  Put  each  number  in  the  form  of  its  factors  (modulus 
and  directing  factor) . 

Then  the  product  is  evidently  the  product  of  the  moduli 
into  the  directing  factor  of  the  product,  whose  angle,  as  for 
points  on  the  unit-circle,  is  the  algebraic  sum  of  the  angles 
of  the  factors.  It  is  necessary  to  use  the  tables,  except  in 
those  few  special  cases  when  the  angles  are  such  that  their 
sines  and  cosines  are  known  without  use  of  the  tables,  as  30°, 
45°,  etc. 


§191] 


THE   FOUR   UNITS. 


351 


(i)   Example  with  tables  not  necessary. 

1  +  i  =  V2(-^  +  —  zj  =  V2(cos  45°  +  i  sin  45°). 

V3  +  i  =  2^—  +|  *)=  2(cos  30°  +  i  sin  30°). 

...  (l  +  0(V3  +  0  =  2  V2(cos  75°  +  t  sin  75°). 

The  sine,  cosine  of  75°  may  now  be  taken  from  the  tables, 
or  expressed  in  radicals. 

(ii)  Example  in  which  tables  are  advisable. 

(375  +  274  0(432  +  548  0. 


375  +  274  i  =  V(375)2  +  (274)2(cos  A  +  i  sin  A), 
where  A  =  tan"1 §£|. 

A  is  best  found  by  the  use  of  logarithms,  and  the  value  of 
the  modulus  from  a  table  of  squares. 

The  second  factor  is  treated  in  the  same  way. 

The  product  is  then  rxr2  J  cos  ( A  +  B)  +  i  sin  (A  +  B)  \ . 

The  student  may  complete  calculations. 

(c)  Pictorial  multiplication. 

Since  the  modulus  of  the  product  of  two  factors  is  the 
product  of  the  moduli  of  the  factors,  and  the  angle  is  the  sum 
of  the  angles,  constructive  mul- 
tiplication is  readily  carried  out 
by  the  use  of  similar  triangles. 
Join  one  of  the  points,  as  Pt 
(Fig.  171),  representing  one  of 
the  factors  of  the  product,  to  the 
unit  point,  A. 

Construct  the  triangle  P20PB 
similar  to  the  triangle  A0Pv  by 
making  the  angle  P20Ps  equal 
to  the  angle  A0Px  and  the 
angle  0P2Ps  equal  to  the  angle 
0APv    P3  represents  the  product  Px 


-P 

& 

Jv 

l\ 

*-  V 

*       V 

-i     -hp+- 

T     '  - 

f     /            p 

Xr'               *dx 

17   t    ^7 

i  j     w  / 

t                t  *'      t 

J-^.*-              4 

(^            7 

O             -A- 

Fig.  171. 


P2,  for  r2 


:1. 


352  PLANE  TRIGONOMETRY.  [§  191 

.«.  r3  =  rxrv  and  the  angle  of  P3  is  the  sum  of  those  of  Pv  P2. 
A  scale  of  equal  parts  to  measure  0P3,  a  protractor  to  meas- 
ure the  angle  AOPz,  and  the  tables,  will  give  an  approximate 
numerical  value  for  P3,  if  it  is  desired.  Groat's  paper  can 
be  used  very  advantageously  when  the  points  are  located  by 
polar  coordinates  or  by  forms  like  r(cos  6  +  i  sin  0). 

EXERCISES. 

1.  Find  the  product  of  (1  4  i)  (  -  1  -  i)  ;  (3  +  4  i)  (5  +  2  i)  ;  (1  +  0 
(-V3  +  0;  (l+V2  +  0(l-V2-i);  (21  +  37  i) (64-18  i)  ;  (273  +  564  i) 
(613  +  515  i). 

2    Simplify  (cos  ^  ~  j  sinfl)10,    (cos  a  +  i  sin  ct)  (cos  /?  +  i  sin  /?) 
(cos  0  +  i  sin  0)12'     (cos  y  -\- i  sin  y)  (cos  8  +  i  sin  8) 
(cos  2  0  -  t  sin  2  fl)7  (cos  3  6  +  i  sin  3  fl)-5 
(cos  4  0  +  i  sin  4  0)12(cos  5  0  -  i  sin  5  6y* 

Hint:  Make  use  of  Ex.  7,  §  190. 

§  192.    Division  of  Points  not  on  the  Unit-circle. 
(a)  Since  (a  +  bi)(a  —  bi)  =  a2  4-  52,  division  by  a  +  fo'  ^s 

multiplication   by  — —  (as  in  case  (a)  of  the  preceding 

section). 

Example:    §±J"     (2  +  3  0(8  ~4  Q  =  ™±i 

3  +  4  i  25  25 

This  is  the  best  method  when  a,  b  are  small. 
(5)  Ify  using  the  directing  angles  of  the  numbers. 
1  1     cos2  0  + sin2  0      1 


r(cos  6  +-  i  sin  0)      r     cos  0  +  i  sin  0      r 


(cos  6  —  i  sin  0) 


=  i(cos  (-  0)  +  t  sin  (-  0)). 

.-.  £-±4?  =  ^  *  (cos  *i  +  *  sin  ^i)(cos  (-  02)  4-  i  sin(-  02)) 
c  4-  di      r2 

=  ^{008(^-^)4^8^(^-^)1, 

where  rv  r2,  are  the  moduli  of  a  4  fo\  c  +  di;  6V  02,  their 
directing  angles. 


§  192]  THE  FOUR  UNITS.  353 

Thus  the  modulus  of  the  quotient  is  the  quotient  of  the 
moduli,  and  the  directing  angle  of  the  quotient  is  that  of 
the  dividend  minus  that  of  the  divisor. 

Two  cases  may  arise  : 

(1)  The  directing  angles  may  be  apparent  without  the  use 
of  tables,  as  in 

V3 


(?+!) 


V3  +  t__        V  2       2  J  2      cos  30°  +  i  sin  30° 

1  +  i  ~  A/g/  1         1    A"  Vl  '  cos 45°  +  t  sin  45° 


if— +— <1 

W2   V2 ; 


=  V2  •  (cos  15°  - 1  sin  15°), 

and  the  final  numerical  value  can  here  be  obtained  either 
with  or  without  the  use  of  tables. 

Such  examples  are  of  infrequent  occurrence,  when  not 
made  to  order. 

(2)   Tables  advisable. 

213  +  315  %  =  V(213)2  +  (315)2     cos  x  +  i  sin  x 
314  +  426  i      V(314)2  +  (426)2     cos#  +  isiny' 

where  tan  x  =  f if ,  and  tan  y  =  |||. 

x,  y  may  be  found  from  tables  and  the  work  carried  out 
as  above. 

Unless  a  table  of  squares  is  at  hand  to  obtain  the  values 
of  the  radicals,  there  is  no  advantage  in  this  method  over 
the  direct  process  of  (a). 

(c)  Pictorial  division. 

This  is  carried  out  similarly  to  pictorial  multiplication. 
Instead  of  advancing  the  terminal  of  the  dividend  let  it  retro- 
grade (algebraically)  by  the  angle  of  the  divisor.  The  tri- 
angle PsOP2  (Fig.  172)  is  made  similar  to  IOPv  by  turning 
OP2  back  (algebraically)  an  angle  equal  to  IOPv  and  making 

at  P2  the  angle  OP2P3  equal  to  IP1 0  (if  rectangular  paper  is 

p 
used).     Then  P3  represents  -^,  for  the  angle  of  Ps  is  that  of 

i  r        1 

P2  minus  that  of  Pv  and,  from  similar  triangles,  —  =  — ,  or 


354 


PLANE  TRIGONOMETRY. 


[§192 


r3  =  -^.     In  the  diagram  P2  is  (14,  50°),  Px  is  (7,  15°),  I 

ri 
(the  unit-point)  is  (5,  0°);  so  P3  is  (10,  35°). 


Fig.  172. 


EXERCISES. 

1.  Divide  1  4-  i  by  —  1  +  V— 3  by  the  first  and  second  methods, 
without  using  the  tables.     Carry  out  the  division  also  pictorially. 

2.  Divide  1+Vdi  bj  -1-V3i,  as  in  Ex.  1. 

3.  Divide  213  +  321  i  by  324  -  245  i. 

4.  Find  the  quotient  of  any  point  by  its  opposite  point  (the  point 
symmetric  to  the  given  point  with  reference  to  the  origin).  What  is 
the  result  when  the  points  are  on  the  unit-circle  ?  Find  also  the  product 
of  such  pairs  of  points. 

5.  Carry  out  pictorial  division  on  pairs  of  points  selected  with  variety 
of  quadrantal  position,  using  Groat's  paper.     Pictorial  multiplication. 


§  193]  THE  FOUR  UNITS.  355 

§  193.    Integral  Roots  of  Numbers  (Points)  on  the  Unit-circle. 

If  Pv  _P2,  P3,  three  points  on  the  unit-circle  whose  angles 

are  4  4  +  120°>   4  +  240°,  be  cubed>  there  will  result  the 

single  point  whose  angle  is  A.  Thus,  there  are  three  cube 
roots  for  any  point  on  the  unit-circle,  and  they  are  120° 
apart  at  the  vertices  of  an  equilateral  triangle. 

It  is  also  easily  apparent  that  there  are  only  three  cube 
roots,  for  if  the  angle  A  locates  a  terminal,  the  same  terminal 
is  located  by  A  +  n  •  360°,  and  since  any  number  divided  by 
three  can  leave  only  the  remainders  0,  1,  2,  the  integer  n 
above  can  have  only  the  three  forms  3m,  3 m 4-  1,  3m  +  2, 

where  m  is  an  integer.     Thus  — ^^ will  be 

4  +  ™<'  360,  or  4  +  m  •  360°  + 120°,  or  4  +  m  •  360°  +  240°. 
o  o  3 

These  angles  locate  only  the  three  terminals  located  by 

4,4  +  120°>  4  +  240°>  as  above.  Thus  there  are  three>  and 
3    3  o 

only  three,  cube  roots  for  a  point  on  the  unit-circle. 

Similarly,  if  the  four  points  on  the  unit-circle  with  the 

angles  4  4  +  90°>  4  + 180°>  4  +  2™°   are   raised   to   the 
4     4  4  4 

fourth  power,  there  will  result  the  single  point  whose  angle 
is  A.  We  have  thus  four,  and  only  four,  fourth  roots  for 
the  point  whose  angle  is  A. 

And,  in  general,  the  wth  power  of  points  whose  angles  are 

A    A     360°    A     0    360°         A      ,       ^360°     ,  . 

_,    _  + ,    _  +  2 ,  .-  -  +  (n-l) ,    («) 

n      n        n        n  n  7iv  n 

will  be  the  single  point  whose  angle  is  A.  And  these  points 
are  the  n  nth  roots  of  the  point  whose  angle  is  A.  The  argu- 
ment for  only  three  cube  roots  holds  for  only  n  nth  roots,  for 

an  nth  of  the  angle  locating  a  given  terminal,  or, , 

can  have  only  the  n  values  (a)  above  plus  multiples  of  360°, 
locating  only  n  different  terminals,  since  r  divided  by  n  can 
leave  only  n  different  remainders. 


356 


PLANE   TRIGONOMETRY. 


[§194 


§  194.   Integral  Roots  of  Positive  Unity  (or  Solution  of 
xn  -  1  =  0). 

This  is  the  special  case  of  the  preceding  paragraph  when 
^  =  0°. 

Thus,  the  three  cube  roots  of  unity  are  located  by  0°,  120°, 
240°,  and  are  1,  cos  120°  +  i  sin  120°,  cos  240°  +  i  sin  240°, 
1         1  .  VS.        1      V3.      rm<     17o  x 

The  four  fourth  roots  of  unity  are  located  by  0°,  90°,  180°, 
270°,  and  are  1,  t,  -  1,  -  u     (Fig.  174.) 

The  five  fifth  roots  of  unity  are  located  by  0°,  72°,  144°, 
216°,  288°,  432°,  and  are  1,  cos  72°  +  *  sin  72°,  etc.,  and  may 
be  calculated  by  the  tables.     (Fig.  175.) 


in  7f 

•k""         ""*>« 

/4£*.            X 

y2*  ^*  V 

1    4    V    -a.    X 

4    -5          - 4 

Iw       A 

X  tzt  zz  l 

\4?-'         -/ 

122-     y 

2^ — ' 

±-££S* 

i. 

I 

^■?Ps 

j7'    **~    "V^k. 

T   i                x^ 

~~l  -*~               '-  K 

-itA                 ii 

\z         -A 

\  ^         ZCt 

\  *z     Sj 

i^lZJL*- 

^iz:^ 

7 

<"T~T' 

J?             f    "v 

-,^     1     ^ 

/    %    jL*_    v 

i    N££    ■- 

i    J?v 

J    ^       A       -/ 

§<f     Y  2 

"* — ** 

Fig.  173. 


Fig.  174. 


Fig.  175. 


The  corresponding  roots  are  indicated  on  Figs.  173,  174, 
and  175,  at  the  corners  of  an  equilateral  triangle  ;  at  the  cor- 
ners of  a  square  ;  at  the  corners  of  a  regular  pentagon,  etc. 

Similarly,  if  n  is  a  positive  integer,  the  n  nth.  roots  of  unity 
are  at  the  corners  of  a  regular  w-gon,  one  of  whose  vertices 
is  at  unity. 

EXERCISES. 

1.  Prove  by  direct  multiplication  that  the  cube  roots  of  1  given  above 
produce  1  when  cubed.     Treat  the  fourth  roots  in  the  same  way. 

2.  Show  by  pictorial  multiplication  that  each  cube  root  above  is  a 
cube  root  of  1.     Treat  the  fourth  and  fifth  roots  the  same  way,  pictorially. 

3.  Find  pictorially  the  six  sixth  roots  of  unity  and  their  values  from 
the  diagram,  and  show  by  pictorial  multiplication  that  each  is  a  sixth 
root  of  unity. 

4.  Use  the  tables  to  calculate  the  7th,  8th,  9th,  10th  roots  of  1. 


§195] 


THE   FOUR  UNITS. 


357 


5.  Solve  by  algebraic  processes  the  equations,  z3  —  1  =  0,  xA  —  1  =  0, 
x*  —  1  =  0,  and  compare  the  results  and  labor  with  the  pictorial  method 
of  solution. 

6.  Show  pictorially  and  algebraically  that  if  a  is  any  complex  cube 
root  of  unity,  a2  +  a  -f  1  =  0. 

7.  Show  pictorially  and  algebraically  that  the  sum  of  the  cube  roots 
of  1  is  zero.  Also  that  the  sum  of  the  fourth  roots  is  zero ;  sum  of  the 
fifth  roots  is  zero ;  sum  of  the  nth  roots  is  zero. 

8.  Show  that  if  a  is  a  complex  fourth  root  of  unity,  «3 + a2 + a  + 1  =  0. 

9.  State  and  prove  the  general  proposition  corresponding  to  Exs.  6 
and  8. 

10.  Show  that  the  sum  of  the  products  of  the  cube  roots  of  unity, 
taken  two  and  two,  is  zero.  Show  this  pictorially  and  algebraically. 
Show  that  the  product  of  the  cube  roots  of  unity  is  +  1. 

11.  Show  for  the  fourth  roots  of  unity  that  the  sums  of  their  products, 
two  at  a  time,  three  at  a  time,  are  zero.  Prove  that  their  product  is  —  1. 
Prove  these  propositions  pictorially  and  algebraically. 

12.  If  you  have  some  acquaintance  with  the  theory  of  equations,  state 
and  prove  the  general  propositions  of  which  Exs.  10  and  11  are  special 


13.   Use  the  tables  to  solve  the  equations: 
xz  =  cos  15°  +  i  sin  15°,  x4  =  cos  24°  +  i  sin  24°,  x6  -  cos  25°  +  i  sin  25°, 
x6  =  cos  144°  +  i  sin  144°,  x8  =  213  +  185°  i.     Give  diagrams. 


§  195.    Integral  Roots  of  Negative  Unity  (or  Solution  of 
xn  +  1  =  0). 

This  is  a  special  case  of  §  193  when 
A  —  180°.  Thus,  the  three  cube  roots  of 
-  1  are  located  by  60°,  60°  +  120°,  60° 

+  240°,  and  are  \  +  ^t,  -  1,  \  -  ^i, 

as  shown  by  Fig.  176. 

The  four  fourth  roots 

of  —  1  are  located  by 

45°,    45°+  90°,     45°  + 


::x 

-/    -     k 

A,                   7^ 

1     S        -Z      % 

r::rs:;z      j_ 

2S" 

\                          -^         -V 

\        7               ^     t 

S^7     -                  ^U 

%                       *t 

z 

Fig.  176. 


180°,  45°+  270°,  and  have  the  four  values 

as 


Fig.  177. 


±  cos  45°  ±  i  sin  45°,    or  ±  — -  ±  — ~  i> 
shown  by  Fig.  177.  V2     V2 


358  PLANE   TRIGONOMETRY.  [§  195 

Similarly,  the  n  nth  roots  of  —  1  are  the  n  values  which 

cos  2r  +  1 180°  +  i  sin  2r  +  1 180°  can  take  when  r  is  0, 1,  2, 

n  n 

3,  "-n  —  1,  these  being  the  only  values  such  an  expression 
can  take  when  r  takes  all  integral  values  from  —  oo  to  +  oo. 

EXERCISES. 

1.  Solve  by  algebraic  process  the  equations  x3  +  1  =  0,  x*  +  1  =  0,  and 
compare  the  results  with  the  pictorial  solution. 

2.  Solve  pictorially  the  equations  x2  +  1  =0,  x5  +  1  =  0,  x6  +  1  =  0, 
x1  +  1  =  0,  x8  +  1  =  0,  using  the  tables  when  necessary  to  get  the  numeri- 
cal solution. 

3.  Show  pictorially  that  each  root  found  in  the  preceding  cases  is  the 
appropriate  root. 

4.  State  and  prove  for  roots  of  negative  unity  propositions  correspond- 
ing to  Exs.  4-8  inclusive,  of  the  preceding  section. 

§  196.     Integral  Roots  of  Any  Number,  a  +  bi. 

Set  the  given  number  in  the  form  of  modulus  times  direct- 
ing factor,  the  latter  being  taken  in  its  general  form.  Then 
an  nth.  root  is  the  ordinary  nth  root  of  the  modulus  times 
any  one  of  the  n  nth  roots  of  the  directing  factor,  found  as 
already  given  for  points  on  the  unit-circle. 

a  +  bi  =  VW+V*  \  cos  (2  r  •  180°  +  A)  -f  *  sin  (2  r  - 180°  +  A)\, 

where  A  is  the  smallest  positive  angle  locating  a  -h  bi, 

a\,2,i2^       2r-180°  +  ^      ••    2r -180° +  A\n 

and  \  (a2  +b  2)2«  cos ~ p  %  sin ! \ 

[  .  n  n  } 

=  a  4-  bi. 

The  n  nth  roots  of  a  +  bi  are  the  n  different  values  which 

the  bracketed   expression   can   take   when   r  is   given   the 

values  0,  1,  2,  3,  •••  n  —  1. 

Thus  the  roots  for  points  on  any  circle  with  radius  R  are 
located  on  the  same  terminal  lines  as  the  corresponding  roots 
for  corresponding  points  on  the  unit-circle,  and  all  lie  on  the 
circle  whose  radius  is  the  ordinary  required  root  of  R. 

The  cube  roots  of  8  are  located  on  the  circle  of  radius  2, 
as  are  the  cube  roots  of  unity  on  the  unit-circle. 


§197]  THE   FOUR  UNITS.  359 

EXERCISES. 

1.  Locate  the  three  cube  roots  of  —  8 ;  the  four  fourth  roots  of  16  and 
—  16 ;  the  five  fifth  roots  of  32  and  —  32.  Give  the  numerical  values  of 
these  roots. 

2.  Use  the  tables  to  calculate  the  six  sixth  roots  of  64  (cos  234°  +  i  sin 
234°) ;  the  three  cube  roots  of  213  +  432  i. 

§  197.     De  Moivre's  Theorem  in  General. 

The  multiplications,  divisions,  powers,  and  roots,  given  in 
the  preceding  paragraphs  are  all  special  cases  of  what  is 
known  as  De  Moivre's  theorem,  which  has  two  forms : 

(a)  (cos  a  -f-  i  sin  a)  (cos  /3  +  i  sin  /3)  (cos  7  -f  i  sin  7)  (  ) 
()•••  =  cos(«  +/3  +  7  +  •••)+*  sin O  +  £+  7+--). 

(5)  (cos  0  -f-  i  sin  6)n  =  cos  nQ  -f-  i  sin  n6. 

Form  (a)  is  only  the  multiplication  of  complex  numbers 
on  the  unit-circle  (§  190). 

Form  (6)  can  be  proven  readily  for  all  real  values  of  w,  — 
integers,  fractions,  irrational  numbers,  —  positive,  negative. 
We  shall  not  consider  the  case  when  n  is  itself  complex. 

(1)  n  a  positive  integer.  The  proof  is  the  generalization 
of  the  multiplication  in  §  190,  for  n  factors  all  equal. 

(2)  n  a  negative  integer. 

Let  n  =  —  m,  so  that  m  is  a  positive  integer. 

Then 

1 


(cos  0  +  i  sin  0)n  =  (cos  6  +  i  sin  0)-™  = 


(cos  6  +  i  sin  0)m 
cos2  m6  -f-  sin2  mQ 


cos  mO  +  i  sin  m  0     cos  m0  +  i  sin  m6 
=  cos  m6  —  i  sin  m6  =  cos  (  —  rn)6  +  i  sin  (  —  ni)  0 
=  cos  n6  -J-  i  sin  nd. 

(3)  n  a  positive  common  fraction  with  numerator  1,  as 

1 
n=  — 

9. 

In  this  case,  instead  of  writing 

\  0  6 

(cos  0  -f  i  sin  0)?  =  cos  -  +  i  sin  -, 


360  PLANE   TRIGONOMETRY.  [§  197 

the  general  angle  2  rir  +  6  locating  the  same  terminal  that 
6  does  should  be  taken,  since  in  this  case  (cos  6  +  i  sin  6y  is 

A  A 

^-valued,  while  cos  -  -f-  i  sin  -  is  single-valued. 

By  (1),  *  » 

(cos  2r7r  +  6  +  i  sin  2r7r  +  <9Y  =  cos  (2  rvr  +  0)  +  e  sin  (2  rir  +  6) 

=  cos  6  +  »  sin  0. 

2nr  +  0      .  .    2r7r  +  i9  .   .        i 

.-.  cos h  z  sm =  (cos  v  +  i  sin  0)'7. 

The  left-hand  member  can  take  q,  and  only  <?,  values,  namely, 
those  it  takes  when  r  =  0,  1,  2  •••  n  —  1. 

(4)  w  a  negative  common  fraction  with  numerator  1 . 
The  proof  follows  from  (3)  as  (2)  from  (1). 

The  result  may  be  set  in  the  form : 

2<7rr  +  <9      .   .    2irr  +  0      ,        ...    ^-| 

cos I  sin =  (cos  6  +  i  sm  0)  ?. 

q  q 

V 

(5)  w  any  positive  common  fraction,  as  — 

Here  also  the  ^th  root  is  £- valued  and  the  general  angle  is 
to  be  taken. 

By  (l), 

2  Trr  +  p0       .   •    2  irr  +  p0Y? 
cos —  +  z  sin £- 

=  cos  (2  r-7r  +  p6)  +  t  sin  (2  rir  +  p#) 
=  cos^>0-M'sin|?# 
=  (cos  6  +  i  sin  fl)? 

.*.  cos —  +  *  sin ±—  =  (cos  6  +  i  sm  0)?. 

The  ^-values  which  the  left-hand  member  can  take  when 

r  =  0,  1,  2,   3  •••#  —  1,  are  the  #- values  of  the  expression 

p 
(cos  0  -H  sin  0)*. 

(6)  W  a  negative  common  fraction. 

The  proof  follows  from  (5)  as  (2)  from  (1). 


(• 


§  197]  THE   FOUR   UNITS.  361 

(7)  n  an  irrational  number. 

An  irrational  number  like  3.14159  •••,  can  be  looked  upon 
as  the  limit  toward  which  the  numbers  3,  3.1,  3.14,  3.141, 
3.1415,  etc.,  are  approaching.  Thus,  by  taking  more  and 
more  figures  of  the  irrational  number  a  common  fraction 
can  be  found  from  which  the  irrational  number  differs  by  an 
amount  as  small  as  we  please.  Between  two  such  fractions, 
one  above  the  irrational  number  and  one  below  it,  lies 
any  irrational  number.  The  theorem  holds  for  the  common 
fractions,  and  thus  in  the  limit  may  be  proved  to  hold  for 
the  irrational  number. 

EXERCISES. 
(A  "cyclic  group.") 


Prove  by  direct  multiplication,  and  also  on  Groat's  polar  coordinate 
paper,  the  following : 

1.   Given  the  six  roots  of  x6  =  1 :  1 ;  cos  -  +  i  sin  -  ;  cos  —  +  i  sin 


?;  cos^  +  isin^; 
3  3  3  3 


cos  7r  +  i  sin  it  ;  cos  -£  +  i  sin  -^j  cos  -^  +  i  sin  -^  • 

O  o  o  o 

(a)  Show  that  the  product  of  any  two  is  a  third  one  of  the  set. 

(b)  Show  that  ( cos  -  +  i  sin  ^ )  where  x  =  0,  1,  2,  3,  4,  5  are  the  six 

given  numbers.     Show  that  the  same  is  true  of  cos h  i  sin Is  it 

true  of  any  other  one  of  the  set  ? 

(c)  To  what  power  must  each  be  raised  to  produce  1  ? 

(d)  Show    that     (cos  f  +  s  sin  -)~l  =  cos  —  +  •  sin  — ,    and    that 

\       o  o  /  3  3 

( cos  —  +  i  sin  — )     =  cos  -  +  i  sin  -  •     Is  this  true  of  any  other  two  ? 

V         o  o  I  o  3 

What  ones  produce  themselves  when  raised  to  the  power  -1  ? 

(e)  Show  that  all  the  six  numbers  can  be  obtained  from  cos  tt  +  i  sin  7r 

and  cos  ^  +  i  sin  ^  by  forming  their  powers  and  successive  products. 

o  o 

Is  this  true  of  any  other  two  of  the  set  ? 

(/)  The  above  numbers  are  an  example  of  a  cyclic  group. 

2.  Write  down  the  roots  of  a;10  =  1,  and  of  x12  =  1  and  discuss  them 
as  the  above  numbers  are  discussed. 

3.  Assuming   De   Moivre's  Theorem,  prove  that  sin  (a  +  /?  +  y)  = 
sin  a  cos  ft  cos  y  +  sin  ft  cos  y  cos  a  +  sin  y  cos  a  cos  /?  —  sin  a  sin  ft  sin  y. 


362 


PLANE   TRIGONOMETRY. 


[§198 


§  198.    Use  of  De  Moivre's  Theorem  to  Express  Cosines  and  Sines 
of  Multiples  of  an  Angle  in  Terms  of  Powers  of  Cosines  and 
Sines  of  the  Angle. 
If  a  +  bi=  c  +  di, 

then  a  =  c,  and  b  =  d, 

for  each  complex  number  locates  but  a  single  point  on  the 

Argand  diagram.     By  (6),  §  197, 

cos  2  0  +  i  sin  2  0  =  (cos  0  +  i  sin  0)2 

=  cos2  0  —  sin2  0  +  i  •  2  sin  0  cos  0. 
Equating  reals  on  opposite  sides,  also  imaginaries  on  oppo- 
site sides  :  Cos  2  0  =  cos2  (9  -  sin2  0. 

sin  2  0=2  sin  0  cos  0,  as  in  §  139. 
The  general  process  is  illustrated  in  this  special  example. 
Similarly : 

cos  3  0  +  i  sin  3  0=  (cos  0  + 1  sin  0)3 

=  cos3  0  -  3  cos  (9  sin2  0 + 1  (3  cos2  0 
sin0— sin30). 
•\  cos  3  0  =  cos3  0  —  3  cos  0  sin2  0 ; 

sin  3  0  =  3  cos2  0  sin  0  -  sin3  0.     (§  144) 

EXERCISES. 

1.  Express  similarly :    cos  4  6,  sin  4  0,  cos  5  0,  sin  5  0,  cos  6  6,  sin  6  0, 
cosn0,  sinnd. 

2.  Express  in  terms  of  tan  0,  the  values  of  tan  2  0,  tan  3  0,  tan  4  0, 
tan  5  0,  tan  6  0,  tan  7  0. 


§  199.  Use  of  De  Moivre's  Theorem  to  Express  Powers  of  Cosines, 
Sines  in  Terms  of  Cosines,  Sines  of  Multiple  Angles. 


Let     cos  0  +  i  sin  0  =  z. 
Then  cos  0  —  i  sin  0  =  -, 

25    J 


and 


cos  n0  -f  i  sin  n0  =  zn. 
1 


cos 


—  i  sin  w0  = 


2  cos  0  =  25  +  -, 

25 

2  i  sin  0  =  2 , 

z  J 


(i) 

(2) 


and 


2  COS  W0  =  25"  +  -, 

zn 


2  i  sin  w0 


z7'  — 


4» 


(3) 
(4) 


§  199  c]  THE   FOUR  UNITS.  363 

(1),  with  the  aid  of  (3),  will  express  any  power  of  a  co- 
sine in  terms  of  cosines  of  multiple  angles. 

(2),  with  the  aid  of  (3)  or  (4),  according  as  an  even  or 
odd  power  is  given,  will  express  any  power  of  a  sine  in  terms 
of  cosines  or  sines  of  multiple  angles. 

Thus,  for  cos3  0, 

by  (1),        23  cos3  0  =  (z  +  -)S=  z3  +  3  z2  •  -  +  3  z  (*Y+  1 
\        zj  z  \zj      z3 

=  2  cos  3  e  +  3  •  2  cos  0. 
.-.  cos3  6  =  { {cos  3  0  +  3  cos  0j. 
Similarly,  for  sin3  0, 
by  (2),  23  i3  sin3  6  =  (z  -  Vfm  z3  -  i  -  3  fz  -  *\ 

=  2  i  sin  3  6  -  3  •  2  i  sin  0. 

.-.  -  sin3  d  =  \  (sin  3  d  -  3  sin  0). 
For  sin4  0, 

by  (2),        2n'*.sin^  =  ^-iy=^  +  i)-4^2  +  y  +  6 

=  2  cos  4  6  -  4  •  2  cos  2  0  +  6 
.\  sin4  0=|  (cos  4  6  -  4  cos  2  6  +  3). 

A  little  study  of  the  preceding  examples  will  make  it  pos- 
sible to  write  the  value  for  any  power  of  sine  or  cosine  with- 
out making  the  actual  expansion  by  the  binomial  theorem : 

(a)  The  division  factor  for  cubes  is  4,  or  22;  for  fourth 
powers  it  is  23 ;  for  nth.  powers  it  is  2n_1. 

(5)  For  powers  of  cosines  there  are  only  cosines  in  the 
second  member,  starting  with  the  same  multiple  angle  as  the 
power  given,  dropping  two  each  term,  the  coefficients  being 
those  of  the  first  half  of  the  binomial  expansion. 

(e)  Even  powers  of  sines  turn  into  cosines,  odd  powers 
into  sines,  with  alternation  of  sign,  the  coefficients  being  those 
of  the  first  half  of  the  binomial  expansion.  The  sign  of  the 
first  member  is  determined  by  that  of  the  corresponding 
power  of  i.     Since  i4  =  1,  the  value  of  any  power  of  i  is  i,  i2, 


364  PLANE   TRIGONOMETRY.  [§  199  rf 

z3,  or   1.      Divide    the    given    power   by   4;    sign   is  -f-  for 
remainders  0,  1 ;  — ,  for  remainders  2,  3. 

(d)  The  binomial  coefficients  (the  first  half)  are  formed 
by  Pascal's  rule,  as  indicated  in  the  following  set : 

1,    2 


1, 

3, 

3 

1, 

4, 

6 

1, 

5, 

10, 

10 

1, 

6, 

15, 

20 

1, 

T, 

21, 

35, 

35 

1, 

8, 

28, 

56, 

70, 

etc. 

A  number  under  a  number  in  any  line  is  formed  from  that 
immediately  above  it  by  the  addition  to  the  latter  of  the 
number  immediately  preceding  it. 

From  such  a  table  we  can  write  immediately  the  value  of 
any  power  covered  by  the  table.     For  example  : 

cos8  d  =  -1  (cos  8  6  +  8  cos  6  6  +  28  cos  4  6  +  56  cos  2  6  +  35), 
sin80  =  i(cos80-8cos60+28cos40-56cos2<9  +  35), 

-  sin7  6  =  4(sm  7  6  -  7  sin  5  6  -h  21  sin  3  6  -  35  sin  0). 

26 

EXERCISES. 

1.  Fill  out  the  preceding  table  of  binomial  coefficients  to  the  twelfth 
power,  and  use  it  to  write  down  the  values  of  the  first  twelve  powers  of 
sine  and  cosine. 

2.  Carry  out  the  expansions  direct  for  cos60,  sin60. 

3.  Show  that  an  expression  for  cos50  sin70  can  be  obtained  by  multiply- 
ing together  the  expansions  for  (x2 ]   and  Ix )  ,  giving 

-  211  cos60  sin70  =  sin  12  $  -  2  sin  10  $  -  4  sin  8  0  +  10  sin  6  0 

+  5  sin  4  0  -  20  sin  2  0. 

4.  Show  that  sinw0  cosn0,  for  any  positive  integral  values  of  m,  n,  can 
be  changed  to  multiple  angles,  by  changing  each  factor,  multiplying 
results  together,  and  then  changing  products  of  functions  of  multiple 
angles  to  sums  (differences)  by  §  147. '   Change  sin80  coslo0. 


§200]  THE   FOUK  UNITS.  365 

5.  Expand  x  cos  v  +  y  sin  v  +  iz  to  the  second,  third,  fourth,  fifth 
powers,  and  show  that  in  each  case  the  coefficients  of  the  sines  (cosines) 
of  the  multiple  angles  are  homogeneous  functions  of  x,  y,  z,  and  that  in 
no  case  are  the  coefficients  linearly  connected,  that  is,  connected  in  a  first 
degree  additive  (subtractive)  relation. 


§  200.  Relation  of  Sines,  Cosines  to  the  Exponential  Imaginary. 
We  have  shown  in  §§  8,  156  that 

/y3t  zyO  nA  T^  V&  f^ 

Assuming  that  (3)  holds  for  x  =  i6, 


and 


[2+[4    +  Y    ia+!5 

..). 

.-.  e**  =  cos  0  +  i  sin  0. 

(4) 

e-«0  =  cos  6  -  i  sin  9. 

(5) 

.'.   5 =  COS0. 

(6) 

ei9-e-i9 

2i      =  sm9' 

(?) 

*  "    *V*>t0    i    *>-»0A  -  lan  W* 

(8) 

The  preceding  equations,  (4),  (5),  (6),  (7),  (8),  give  the 
connection  between  the  trigonometric  and  exponential  func- 
tions. 

EXERCISES. 

1.  Use  (6),  (7)  to  prove  the  addition-subtraction  formulas. 

2.  Use  (6),  (7)  to  prove  sin  x  +  sin  y  =  2  sin  x  ^  cos  x  ~  y,  and  the 
other  corresponding  formulas. 


366 


PLANE   TRIGONOMETRY. 


[§201 


§  201.   Periodicity  of  the  Exponential  Function. 

By  equation  (4)  of  §  200, 
eie  =  cos  6  +  *  sin  6  =  cos  (2  nir  -f  0)  +*  sin  (2  mr  +  0)  =  e1'6 


,2«7ri 


1. 


Thus,  exponential  #,(V'),  is  periodic,  the  period  being  2  iri. 

§  202.   Every  Number  has  an  Infinite  Number  of  Logarithms, 
(a)  Base  e.         y  =  ex  =  ^+2™",  by  §  201. 

.-.  a?  +  2  W7rt,  n  taking  all  integer  values  from  minus  in- 
finity to  plus  infinity,  is  the  infinite  set  of  logarithms  of  y, 
x  being  the  ordinary  table-logarithm. 

(6)  Any  base. 

If  ax  =  y  and  elogea  =  a, 

then  y  =  ex  log« a  =  ex  loge a+2ra7ri, 

and  #  loge  a  +  2  mri  makes  up,  as  before,  the  infinite  set  of 
logarithms. 

§  203.    Logarithmic  Picture  of  the  Points  in  a  Plane. 

The  numbers  a  +  hi,  when  a,  b  take  all  real  values,  cover 

a  plane  completely  (§  187).  We 
may  now  take  the  logarithms 
of  all  numbers  in  one  plane  and 
plot  these  logarithms  on  another 
plane. 
Since 


t 

1 

7 

r 

~l      -i 

t     B- 

4 

7 

2ViE 

rj 

Fig.  178.  —  a,  b  plane. 


a  +  bi  =  Va2  +  62(cos  0  +  i  sin  0), 

where  tan  #  =  -,  anv  number  can 

a 

be  set  in  the  form  R  -  eid,  where 
R  is  its  modulus. 

.-.  N=  Reie  =  R-  ei0+2nn\ 
log,  iV=  loge  R  +  i(0  +  2  nir) 
=  x  +  iy. 


204] 


THE   FOUR  UNITS. 


367 


Here  y  is  infinite-valued,  as  n  goes  from 

—  oo    to   -f  00. 

If  N  is  a  point  on  the  unit-circle, 
log  eB  =  0,  and  thus  points  on  the  unit- 
circle  of  one  plane  plot  into  points  on  the 
?/-axis  of  the  other  plane.  Any  point  of 
one  plane  plots  into  an  infinite  number 
of  points,  2  7r  distant  from  each  other,  on 
a  line  parallel  to  the  ?/-axis,  at  a  distance 
x,  or  loge  B  from  it. 

The  point  P  thus  plots  into 

p      p      p     ...    pr      pt     ... 

■*■  V  ■*■  2'  -*•  3'        >  ■*■     V         3'        ' 

where  0M=\ogeB, 

MP1  =  0, 
PXP2  =  2  7T,  etc. 


— pH-^J^4 

1,  Af 

;g'  :  0=*2  ir 

'°L  2k"5  7r 

1~          1 

Fia.  179.  —  x,  y  plane. 


§  204.     Mercator's  Projection. 

If  A,  B  are  the  extremities  of  a  diameter  of  a  sphere,  and 
if  a  plane  touches  the  sphere  at  A,  and  if  a  straight  line  is 

then  drawn  from  B  through  each 
point  of  the  sphere  back  to  the 
plane,  we  have  what  is  called  a 
polar  projection  of  the  sphere 
on  the  plane.  If  A,  B  represent 
the  north  and  south  poles  of 
the  earth,  the  earth's  meridian 
curves  become  rays  from  A  on 
the  plane  and  the  latitude 
curves  become  concentric  cir- 
cles about  A. 

If  now,  we  take  the  logarithm 
of  this  polar  projection,  we  have,  when  these  logarithms  are 
plotted,  Mercator's  Projection.  Which  circle  on  the  polar 
projection  is  taken  for  the  unit-circle  is,  of  course,  quite 
arbitrary.  This  unit-circle  plots  (§  203)  into  the  y-axis. 
By  §  203,  circles  concentric  with  and  outside  the  unit-circle 


X           x 

K~           X 

--*s         +-«= 

^xJ5t  -  vv 

-7       2*          V 

Z.         ^^         5 

4                                        ^                      ^ 

t                                                  ^V            5 

-         f-                                         ^         A 

4.              ^      ^R 

il                                                                                             T3 

X                                                                        J 

\                                                                     / 

X                        7- 

\                                          / 

— --_           ^-^^ 

Fig.  180. 


368 


PLANE   TRIGONOMETRY. 


[§204 


Fig.  181.  — Polar  projection. 


plot  into  lines  parallel  to  the  y-axis,  and  to  the  right  of  it, 
while  those  within  the  unit-circle  plot  into  lines  to  the  left  of 

the  ?/-axis  and  par- 
allel to  it.  Radial 
lines  on  the  polar 
projection  plot 
into  lines  parallel 
to  the  rr-axis  on 
the  Mercator  map. 
The  radius  whose 
directing  angle  is  a 
radian  goes  a  unit's 
distance  up,  etc. 

Thus,  if  we  take 
the  logarithm  of 
the  numbers  repre- 
sented by  a  sheet 
of  Groat's  polar 
coordinate  paper 
in  radian  measure  (a  picture  of  a  polar  projection  of  the  earth) 
we  have  the  numbers  represented  by  a  sheet  of  rectangular 
coordinate  paper  (a  picture  of 
Mercator's  projection). 

On  a  polar  projection  the 
curve  representing  a  direct  sail 
from  Liverpool  to  New  York 
would  be  represented  by  a  curve 
cutting  all  meridian  circles  (the 
radii)  at  the  same  angle.  This 
curve  is  called  an  equiangu- 
lar spiral.  The  logarithms  of 
points  on  this  spiral  pass  into 
a  straight  line  on  the  Mercator 
map,  namely,  the  line   joining 

the  point  which  is  the  logarithm  of  the  point  representing 
Liverpool  to  that  which  is  the  logarithm  of  the  point  repre- 
senting New  York. 


- 

- 

c 

3 

J 

Fig.  182.  — Mercator's  projection. 


§205]  THE   FOUR  UNITS.  369 

If  log  (a  +  bi)  =  log,  R  +  i(d  +  2  mr) 

any  point  on  the  unit-circle  will  become,  in  the  loga- 
rithmic-picture, an  infinite  number  of  points  on  the  ?/-axis, 
separated  from  each  other  by  a  distance  2  it.  If  we  im- 
agine a  point  to  start  at  the  initial  line  and  move  around 
the  unit-circle,  how  would  the  points  representing  the 
logarithms  move  ?  Circles  outside  the  unit-circle  and  con- 
centric with  it  will  turn  into  what?  If  R  is  less  than  1, 
log,  R  is  negative ;  thus  a  point  within  the  unit-circle  will 
lie  to  the  left  of  the  y-axis  in  the  logarithmic-picture.  A 
circle  within  the  unit-circle,  and  concentric  with  it,  will 
become  a  line  parallel  to  the  #-axis  and  to  the  left  of  it. 
Since  for  points  on  a  ray  through  the  origin  in  the  a,  b 
picture,  0  is  a  constant,  the  corresponding  points  in  the 
logarithmic-picture  will  lie  on  lines  parallel  to  the  #-axis,  at 
distances  0  +  2  nir  from  it.  Thus  any  such  ray  will  turn 
into  an  infinite  number  of  parallel  lines.  The  part  of  a  ray 
beyond  the  unit-circle  goes  into  what  ?  What  represents  the 
part  of  the  ray  within  the  unit-circle  ? 

§  205.     The  /th  Power  and  /th  Roots  of  Numbers. 

Since  a  +  bi  can  be  put  in  the  form  R  •  eid,  every  number 
can  be  set  in  the  form  of  two  factors,  the  real  factor  giving 
the  distance  of  its  representing  point  from  the  origin  and  the 
exponential  imaginary  factor  indicating  its  angle. 

.-.  (a  +  biy  =  (R  •  eieY=  e~e  •  Rl  =  i  .  eil0*e*. 

Thus,  if  the  ith.  power  of  a  number  on  one  plane  is  plotted 
on  another  plane,  the  new  modulus  is  the  reciprocal  of  the 
exponential  of  the  old  angle,  while  the  new  angle  is  the  loga- 
rithm of  the  old  modulus. 

Thus  the  ith  power  of  all  points  on  the  unit-circle  will  lie 

on  the  initial  line. 

i  ii 

Similarly,       (a  +  biy  =  (R  •  e19)'  =  e9  •  R1 

=  e9  -R-*  =  e9  -e-ilogR. 


370 


PLANE   TRIGONOMETRY. 


[§205 


Example  :  Find  the  ith.  power  of  •'. 
The  angle  of  i  is  —  and  its  modulus  is  1. 


*=* — s£ 

7          /^ 

J          J&    A 

t          ^tivVA 

/°± 

X 

~X       -t    l 

\                t 

-^         z 

^ 2? 

Fig.  183. 


EXERCISE. 

Make  maps  of  the  ith  power  and  ith  roots  of  all  points  on  a  selected 
curve  on  a  plane. 

§  206.    The  Hyperbolic  Functions. 

If  x,  y  are  the  coordinates  of  any  point  on  a  circle  of 
radius  a,  as  in  Fig.  183,  x2-\-y2  =  a2. 

Thus,  if  we  call  -  the  cosine  of  6  and  U. 
a  a 

the  sine  of  0,  then  characteristic  of  the 
trigonometry  of  the  circle  is 
.      cos2  0+  sin2  (9=1. 
Similarly,  if  x,  y  are  connected  by 
the  relation 

x*-y*=a%  or  (ff-  (£f  =  1, 


_    I 

the  point  x,  y  (Fig.  184)  will  lie  on  the  curve  which  has  the 
name  equilateral  hyperbola, 
or  rectangular  hyperbola,  a 
curve  obtained  by  making  a 
sketch  of  points  (in  a  plane) 
such  that  the  distance  of  each 
from  a  fixed  point  in  that 
plane  bears  to  its  distance 
from  a  fixed  line  (vertical) 
of  that  plane  the  ratio  of  the 
hypothenuse  of  a  45°  right- 
angled  triangle  to  a  side  of 
the  same  triangle. 

In   correspondence   to  the 

trigonometry  of  the  circle,  we  may  call  -,  as  related  to  the 

a 


\                                            z 

\                                               7 

\                                7 

\                                / 

\                           J 

3                                 t 

\                         2  - 

v                    -T-v 

X                         1    4-Ih 

A                         t    - 

*- X :> 

«■  -g  -^ 

3C                           t 

t                 X 

■J                                    K- 

L      ±                      5 

J                                     S 

y                                      \ 

_/                                          \ 

_/                                               ^ 

7                                                    \ 

7                                                              \ 

Fig.  184. 


§207]  THE   FOUR   UNITS.  371 

hyperbola,  the   hyperbolic   cosine,  and  V-  the  hyperbolic  sine, 
so  that 

(hyperbolic  cosine)2  —  (hyperbolic  sine)2  =  1. 

There  is  thus  a  trigonometry  related  to  such  an  hyperbola 
as  is  the  ordinary  trigonometry  to  the  circle. 

Abbreviated  names  of  the  functions  in  this  trigonometry  are  the  same 
as  in  those  of  the  circle,  with  an  h  interpolated :  cosh  for  cos ;  sinh  for 
sin ;  sech  for  sec,  etc.  These  names  may  be  read  "  cosine  hyperbolic," 
"  sine  hyperbolic,"  etc.  It  is  becoming  customary,  for  brevity's  sake,  to 
throw  the  h  in  the  pronunciation  where  it  will  make  the  abbreviated 
form  pronounceable.  Thus  cosh  x  is  pronounced  as  here  written, 
"  cosh  "  x.  Sinh  x  is  "  shin  "  x.  Sech  x  is  "  shec  "  x.  Tanh  x  is  "  than  "  x 
(th  having  its  sound  in  thing),  etc. 

gx  _i_  g-w:  eix e-ix 

We  found  cos  x  = and  sin  x  =  — — (§  200) 

A  At 

As  defining  relations  for  cosh  x  and  sinh  x,  the  i  is  omitted, 

pX  _i_  p- 3D  &X  p  —  OC 

and  coshx  = — ^ and  sinh  x  = 

Whence  cosh2  x  —  sinh2  x  =  l, 

corresponding  to  the  hyperbola  —  —  £-  =  1 . 

§  207.   Relation  of  Hyperbolic  Trigonometry  to  Circular 
Trigonometry. 

From  the  exponential  relations  of  §§  200,  206  follow  at 

once  :  cosh x  =  cos  (ix)  (*■) 

i  •  sinh  x  =  sin  (ix).  (2) 

Thus,  assuming  that  circular  trigonometry  holds  for  quan- 
tity of  the  form  (ix),  the  corresponding  hyperbolic  trigo- 
nometry comes  at  once  by  changing  the  word  cosine  to  cosh 
and  the  word  sine  to  t  •  sinh. 

For  example : 

sin  (x + y)  =  sin  x  cos  y  -f-  cos  x  sin  y . 
.*.  i  •  sinh  (x  +  y}  —  (i  .  sinh  a;) (cosh  y)  -f-  (cosh  x)(i  •  sinh  y), 
or,         sinh  (x  -f  y)  =  sinh  x  cosh  y  +  cosh  x  sinh  y. 


372  PLANE   TRIGONOMETRY.  [§207 

A  second  example  is  cos2#  +  sin2  a;  =  1. 
.*.  (cosh  a;)2  +  (i  •  sinhz)2  =  1. 
.-.  cosh2#  —  sinh2#  =  1. 
From  (1),  (2)  above  it  follows  that 
tan  changes  to  i  •  tanh, 
cot  changes  to  —  i-  coth, 
sec  changes  to  sech, 
cosec  changes  to  —  i-  cosech. 

EXERCISES. 

Prove  from  the  defining  relations  for  the  hyperbolic  functions,  or 
from  the  changes  in  circular  functions  noted  above,  as  may  seem  best, 
the  following : 

1.  sinh  2  x  =  2  sinh  x  cosh  x. 

2.  cosh  2  x  =  2  cosh2  x  -  1  =  2  sinh2  x  +  1. 

3.  cosh  (x  +  y)  =  cosh  x  cosh  y  +  sinh  x  sinh  y. 

4.  cosh  (x  +  y)  —  cosh  (x  —  y)  =  2  sinh  x  sinh  y. 

5.  cosh  3  x  =  4  cosh3  #  —  3  cosh  a\ 

6.  sinh  3  x  =  3  sinh  a;  +  4  sinh3  a;. 

7.  tanh<>  +  y)  =  tanh x  +  tanh SS  . 

1  +  tanh  a;  tanh  # 

8.  sinh  (a:  +  y)  cosh  (x  —  ?/)  =  £  (sinh  2  x  +  sinh  2  y). 

9.  tanh2*=     2  tanh  *    • 

1  +  tanh2  a: 

10.  sinh  0  =  0;  cosh  0  =  1. 

11.  sinh  S  —  i ;  cosh  52  =  0 ;  sinh  7ri  =  0 ;  cosh  vi  =  —  1- 

12.  sech2  a;  +  tanh2  x  =  1 ;  coth2  a;  —  cosech2  x  =  1. 

13.  sinh  ar  -  sinh  v  =  2  sinh  ^^  cosh'^tll. 

2  2 

14.  sinh  x  +  sinh  y  =  2  sinh  -  ~*~  -y  cosh 


2  2 

15.  cosh  x  —  cosh  y  =  2  sinh  x  ~  y  sinh  x  +  y. 

*         2       2 

16.  cosh  x  +  cosh  y  =  2  cosh  x  ~  &  cosh  ar  +  ^« 

■  .     *  2  2 

17.  Plot  the  curves  y  =  ex  and  y  —  e~x  and  thus  obtain  graphs  for 
y  =  sinh  a:  and  y  —  cosh  a;.  Show  that  the  latter  has  the  form  taken  by 
a  string  hung  on  two  pegs  in  the  same  horizontal  line  (catenary). 


208]  THE   FOUR  UNITS.  373 


§  208.  Relation  of  the  Anti-Hyperbolic  Functions  to  Logarithms. 

sinh  x  =  ;  cosh  x  =      ^e    ■         (§  206) 

Let  sinh  x  =  y,  so  that  z  =  sinh-1  ?/. 

.-.  e*  -e~x  =  2  y.      .'.  ez  -  \  =  2  y. 

€■ 

.'.  e2x-2yex-l  =  0. 
.-.  ez  =  y  ±-\fy2  4-  1. 
.-.  a?  =  sinh-1  y  =  loge  (y  ±  Vy«  +  l).  (1) 

Similarly,  cosh-1  y  =  loge  (y  ±  Vy2  _  1}.  (2) 

Since  the  quantity  in  the  bracket  after  log  in  (1)  is  nega- 
tive when  the  minus  sign  is  given  the  radical,  and  since 
negative  numbers  have  no  real  logarithms,  it  is  customary, 
when  the  generality  of  the  logarithm  has  not  been  considered, 
to  take  the  sign  of  the  radicals  in  (1),  (2)  as  plus.  Here,  as 
the  student  is  familiar  with  log  (<z  +  bi),  for  all  values  of  #, 
6,  we  need  not  restrict  the  sign  of  the  radical. 

i  sinh  x     ex  —  e~x 

tanh  x  =  — ; —  = =  y, 

cosh  x      ex  -f-  e  x 

,_*^zl  =  y.     ,.>-.l±JL 


e2*  +  1      *  1-y 

•\  x  =  tanh  ~  V  as  -  log  _      M> 
U      2     *l-y 


Thus,  in  general,  making  y 


a  a 


z  ±  V«a  -  a? 


cosh-1  — =  log 

a  a 

tanh-i^log^?. 
a     2       z-a 


EXERCISES. 

1.  Show  that  the  hyperbolic  functions  are  periodic. 

2.  Show  that  the  anti-hyperbolic  functions  are  many-valued. 


374 


PLANE   TRIGONOMETRY. 


[§209 


§  209.   Relation  of  the  Hyperbolic  Functions  to  the  Circular  Func- 
tions of  the  Eccentric  Angle  of  the  Hyperbola. 

For  the  hyperbola,         x2  —  y2  =  a2. 

...  £!_2*-l. 

a2      a2 

In  the  diagram,  MT  being  a  tangent  to  the  circle, 

-  =  sec  <p, 
<f>  being  called  the  eccentric  angle. 


V 


E 


wm 


A 


M 


And  since 
Also 


and 


(1) 

(2) 


Fig.  185. 

y 

sec2$  —  tan2<£  =  1,     .*.  -  =  tan  <j>. 
cosh20-sinh20  =  l. 
•\  cosh  6  =  sec  <j>, 
sinh  6  =  tan  <f>. 
,\  (f>  =  sec-1(cosh  0)  =  tan_1(sinh  0). 
The  angle  <j>  as  defined  by  these  relations  is  called  the 
G-udermannian  of  6.     (1),  (2)  may  be  put  in  the  forms 
cos  <j>  n  sech  0, 

sin  <£  =  cos  <f>  •  sinh  0  =  tanh  0. 
,\  gd  0  =  cos"1  sech  0  ==  sin-1  tanh  6   =  tan-1  sinh  0. 

EXERCISE. 

Show  that  if  u  =  log  tan  f  J  + 1\ 

w  =  gd-1a\ 


14  DAY  USE 

RETURN  TO  DESK  FROM  WHICH  BORROWED 

LOAN  DEPT. 

This  book  is  due  on  the  last  date  stamped  below,  or 

on  the  date  to  which  renewed. 

Renewed  books  are  subject  to  immediate  recall. 


22May'53AJ 


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m^ 


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9Nov'59FK 


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ocm  m 


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W  H  1960 


TOEC?» 


sep  at 


LD  21A-50m-9,'58 
(6889sl0)476B 


S 


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